Check if given Preorder, Inorder and Postorder traversals are of same tree
Given Preorder, Inorder, and Postorder traversals of some tree. Write a program to check if they all are of the same tree.
Examples:
Input: Inorder -> 4 2 5 1 3
Preorder -> 1 2 4 5 3
Postorder -> 4 5 2 3 1
Output: Yes
Explanation: All of the above three traversals are of the same tree
1
/ \
2 3
/ \
4 5
Input: Inorder -> 4 2 5 1 3
Preorder -> 1 5 4 2 3
Postorder -> 4 1 2 3 5
Output: No
The most basic approach to solve this problem will be to first construct a tree using two of the three given traversals and then do the third traversal on this constructed tree and compare it with the given traversal. If both of the traversals are the same then print Yes otherwise print No. Here, we use Inorder and Preorder traversals to construct the tree. We may also use Inorder and Postorder traversal instead of Preorder traversal for tree construction. You may refer to this post on how to construct a tree from given Inorder and Preorder traversal. After constructing the tree, we will obtain the Postorder traversal of this tree and compare it with the given Postorder traversal.
Below is the implementation of the above approach:
/* C++ program to check if all three given
traversals are of the same tree */
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct Node
{
int data;
struct Node *left, *right;
};
// Utility function to create a new tree node
Node* newNode(int data)
{
Node *temp = new Node;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
/* Function to find index of value in arr[start...end]
The function assumes that value is present in in[] */
int search(int arr[], int strt, int end, int value)
{
for (int i = strt; i <= end; i++)
{
if(arr[i] == value)
return i;
}
}
/* Recursive function to construct binary tree
of size len from Inorder traversal in[] and
Preorder traversal pre[]. Initial values
of inStrt and inEnd should be 0 and len -1.
The function doesn't do any error checking for
cases where inorder and preorder do not form a
tree */
Node* buildTree(int in[], int pre[], int inStrt,
int inEnd)
{
static int preIndex = 0;
if(inStrt > inEnd)
return NULL;
/* Pick current node from Preorder traversal
using preIndex and increment preIndex */
Node *tNode = newNode(pre[preIndex++]);
/* If this node has no children then return */
if (inStrt == inEnd)
return tNode;
/* Else find the index of this node in
Inorder traversal */
int inIndex = search(in, inStrt, inEnd, tNode->data);
/* Using index in Inorder traversal,
construct left and right subtress */
tNode->left = buildTree(in, pre, inStrt, inIndex-1);
tNode->right = buildTree(in, pre, inIndex+1, inEnd);
return tNode;
}
/* function to compare Postorder traversal
on constructed tree and given Postorder */
int checkPostorder(Node* node, int postOrder[], int index)
{
if (node == NULL)
return index;
/* first recur on left child */
index = checkPostorder(node->left,postOrder,index);
/* now recur on right child */
index = checkPostorder(node->right,postOrder,index);
/* Compare if data at current index in
both Postorder traversals are same */
if (node->data == postOrder[index])
index++;
else
return -1;
return index;
}
// Driver program to test above functions
int main()
{
int inOrder[] = {4, 2, 5, 1, 3};
int preOrder[] = {1, 2, 4, 5, 3};
int postOrder[] = {4, 5, 2, 3, 1};
int len = sizeof(inOrder)/sizeof(inOrder[0]);
// build tree from given
// Inorder and Preorder traversals
Node *root = buildTree(inOrder, preOrder, 0, len - 1);
// compare postorder traversal on constructed
// tree with given Postorder traversal
int index = checkPostorder(root,postOrder,0);
// If both postorder traversals are same
if (index == len)
cout << "Yes";
else
cout << "No";
return 0;
}
/* Java program to check if all three given
traversals are of the same tree */
import java.util.*;
class GfG {
static int preIndex = 0;
// A Binary Tree Node
static class Node
{
int data;
Node left, right;
}
// Utility function to create a new tree node
static Node newNode(int data)
{
Node temp = new Node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
/* Function to find index of value in arr[start...end]
The function assumes that value is present in in[] */
static int search(int arr[], int strt, int end, int value)
{
for (int i = strt; i <= end; i++)
{
if(arr[i] == value)
return i;
}
return -1;
}
/* Recursive function to construct binary tree
of size len from Inorder traversal in[] and
Preorder traversal pre[]. Initial values
of inStrt and inEnd should be 0 and len -1.
The function doesn't do any error checking for
cases where inorder and preorder do not form a
tree */
static Node buildTree(int in[], int pre[], int inStrt, int inEnd)
{
if(inStrt > inEnd)
return null;
/* Pick current node from Preorder traversal
using preIndex and increment preIndex */
Node tNode = newNode(pre[preIndex++]);
/* If this node has no children then return */
if (inStrt == inEnd)
return tNode;
/* Else find the index of this node in
Inorder traversal */
int inIndex = search(in, inStrt, inEnd, tNode.data);
/* Using index in Inorder traversal,
construct left and right subtress */
tNode.left = buildTree(in, pre, inStrt, inIndex-1);
tNode.right = buildTree(in, pre, inIndex+1, inEnd);
return tNode;
}
/* function to compare Postorder traversal
on constructed tree and given Postorder */
static int checkPostorder(Node node, int postOrder[], int index)
{
if (node == null)
return index;
/* first recur on left child */
index = checkPostorder(node.left,postOrder,index);
if(index == -1) return index;
/* now recur on right child */
index = checkPostorder(node.right,postOrder,index);
if(index == -1) return index;
/* Compare if data at current index in
both Postorder traversals are same */
if (node.data == postOrder[index])
index++;
else
return -1;
return index;
}
// Driver program to test above functions
public static void main(String[] args)
{
int inOrder[] = {4, 2, 5, 1, 3};
int preOrder[] = {1, 2, 4, 5, 3};
int postOrder[] = {4, 5, 2, 3, 1};
int len = inOrder.length;
// build tree from given
// Inorder and Preorder traversals
Node root = buildTree(inOrder, preOrder, 0, len - 1);
// compare postorder traversal on constructed
// tree with given Postorder traversal
int index = checkPostorder(root,postOrder,0);
// If both postorder traversals are same
if (index == len)
System.out.println("Yes");
else
System.out.println("No");
}
}
# Python3 program to check if
# all three given traversals
# are of the same tree
class node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
preIndex = 0
# Function to find index of value
# in arr[start...end]. The function
# assumes that value is present in in
def search(arr, strt, end, value):
for i in range(strt, end + 1):
if(arr[i] == value):
return i
# Recursive function to construct
# binary tree of size lenn from
# Inorder traversal in and Preorder
# traversal pre[]. Initial values
# of inStrt and inEnd should be 0
# and lenn -1. The function doesn't
# do any error checking for cases
# where inorder and preorder do not
# form a tree
def buildTree(inn, pre, inStrt, inEnd):
global preIndex
if(inStrt > inEnd):
return None
# Pick current node from Preorder
# traversal using preIndex and
# increment preIndex
tNode = node(pre[preIndex])
preIndex += 1
# If this node has no children
# then return
if (inStrt == inEnd):
return tNode
# Else find the index of this
# node in Inorder traversal
inIndex = search(inn, inStrt,
inEnd, tNode.data)
# Using index in Inorder traversal,
# construct left and right subtress
tNode.left = buildTree(inn, pre, inStrt,
inIndex - 1)
tNode.right = buildTree(inn, pre,
inIndex + 1, inEnd)
return tNode
# function to compare Postorder traversal
# on constructed tree and given Postorder
def checkPostorder(node, postOrder, index):
if (node == None):
return index
# first recur on left child
index = checkPostorder(node.left,
postOrder,
index)
# now recur on right child
index = checkPostorder(node.right,
postOrder,
index)
# Compare if data at current index in
# both Postorder traversals are same
if (node.data == postOrder[index]):
index += 1
else:
return - 1
return index
# Driver code
if __name__ == '__main__':
inOrder = [4, 2, 5, 1, 3]
preOrder = [1, 2, 4, 5, 3]
postOrder = [4, 5, 2, 3, 1]
lenn = len(inOrder)
# build tree from given
# Inorder and Preorder traversals
root = buildTree(inOrder, preOrder,
0, lenn - 1)
# compare postorder traversal on
# constructed tree with given
# Postorder traversal
index = checkPostorder(root, postOrder, 0)
# If both postorder traversals are same
if (index == lenn):
print("Yes")
else:
print("No")
# This code is contributed by Mohit Kumar 29
/* C# program to check if all three given
traversals are of the same tree */
using System;
public class GfG
{
static int preIndex = 0;
// A Binary Tree Node
class Node
{
public int data;
public Node left, right;
}
// Utility function to create a new tree node
static Node newNode(int data)
{
Node temp = new Node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
/* Function to find index of
value in arr[start...end]
The function assumes that
value is present in in[] */
static int search(int []arr, int strt,
int end, int value)
{
for (int i = strt; i <= end; i++)
{
if(arr[i] == value)
return i;
}
return -1;
}
/* Recursive function to construct
binary tree of size len from Inorder
traversal in[] and Preorder traversal
pre[]. Initial values of inStrt and
inEnd should be 0 and len -1. The
function doesn't do any error checking for
cases where inorder and preorder do not form a
tree */
static Node buildTree(int []In, int []pre,
int inStrt, int inEnd)
{
if(inStrt > inEnd)
return null;
/* Pick current node from Preorder traversal
using preIndex and increment preIndex */
Node tNode = newNode(pre[preIndex++]);
/* If this node has no children then return */
if (inStrt == inEnd)
return tNode;
/* Else find the index of this node in
Inorder traversal */
int inIndex = search(In, inStrt, inEnd, tNode.data);
/* Using index in Inorder traversal,
construct left and right subtress */
tNode.left = buildTree(In, pre, inStrt, inIndex - 1);
tNode.right = buildTree(In, pre, inIndex + 1, inEnd);
return tNode;
}
/* function to compare Postorder traversal
on constructed tree and given Postorder */
static int checkPostorder(Node node, int []postOrder, int index)
{
if (node == null)
return index;
/* first recur on left child */
index = checkPostorder(node.left,postOrder,index);
/* now recur on right child */
index = checkPostorder(node.right,postOrder,index);
/* Compare if data at current index in
both Postorder traversals are same */
if (node.data == postOrder[index])
index++;
else
return -1;
return index;
}
// Driver code
public static void Main()
{
int []inOrder = {4, 2, 5, 1, 3};
int []preOrder = {1, 2, 4, 5, 3};
int []postOrder = {4, 5, 2, 3, 1};
int len = inOrder.Length;
// build tree from given
// Inorder and Preorder traversals
Node root = buildTree(inOrder, preOrder, 0, len - 1);
// compare postorder traversal on constructed
// tree with given Postorder traversal
int index = checkPostorder(root, postOrder, 0);
// If both postorder traversals are same
if (index == len)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
/* This code is contributed PrinciRaj1992 */
<script>
/* Javascript program to check if all three given
traversals are of the same tree */
let preIndex = 0;
// A Binary Tree Node
class Node
{
// Utility function to create a new tree node
constructor(data)
{
this.data=data;
this.left=this.right=null;
}
}
/* Function to find index of value in arr[start...end]
The function assumes that value is present in in[] */
function search(arr,strt,end,value)
{
for (let i = strt; i <= end; i++)
{
if(arr[i] == value)
return i;
}
return -1;
}
/* Recursive function to construct binary tree
of size len from Inorder traversal in[] and
Preorder traversal pre[]. Initial values
of inStrt and inEnd should be 0 and len -1.
The function doesn't do any error checking for
cases where inorder and preorder do not form a
tree */
function buildTree(In,pre,inStrt,inEnd)
{
if(inStrt > inEnd)
return null;
/* Pick current node from Preorder traversal
using preIndex and increment preIndex */
let tNode = new Node(pre[preIndex++]);
/* If this node has no children then return */
if (inStrt == inEnd)
return tNode;
/* Else find the index of this node in
Inorder traversal */
let inIndex = search(In, inStrt, inEnd, tNode.data);
/* Using index in Inorder traversal,
construct left and right subtress */
tNode.left = buildTree(In, pre, inStrt, inIndex-1);
tNode.right = buildTree(In, pre, inIndex+1, inEnd);
return tNode;
}
/* function to compare Postorder traversal
on constructed tree and given Postorder */
function checkPostorder(node,postOrder,index)
{
if (node == null)
return index;
/* first recur on left child */
index = checkPostorder(node.left,postOrder,index);
/* now recur on right child */
index = checkPostorder(node.right,postOrder,index);
/* Compare if data at current index in
both Postorder traversals are same */
if (node.data == postOrder[index])
index++;
else
return -1;
return index;
}
// Driver program to test above functions
let inOrder=[4, 2, 5, 1, 3];
let preOrder=[1, 2, 4, 5, 3];
let postOrder=[4, 5, 2, 3, 1];
let len = inOrder.length;
// build tree from given
// Inorder and Preorder traversals
let root = buildTree(inOrder, preOrder, 0, len - 1);
// compare postorder traversal on constructed
// tree with given Postorder traversal
let index = checkPostorder(root,postOrder,0);
// If both postorder traversals are same
if (index == len)
document.write("Yes");
else
document.write("No");
// This code is contributed by patel2127
</script>
Output
Yes
Time Complexity : O( n * n ), where n is number of nodes in the tree.
Space Complexity: O(n), for call stack
Efficient algorithm using hash map to store indices of inorder elements :
While building the tree from Inorder and Preorder traversal we need to check if the inorder and preorder traversals are valid themself for some tree, and if yes , then keep building the tree, but if valid binary tree can not be built from given inorder and preorder traversal, then we must stop building the tree and return false. And also we can build the tree from inorder and preorder traversal in O(n) time using hashmap to store the indices of the inorder elements’ array.
Implementation:
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node *left, *right;
Node(int val)
{
data = val;
left = right = NULL;
}
};
Node* buildTreeFromInorderPreorder(
int inStart, int inEnd, int& preIndex, int preorder[],
unordered_map<int, int>& inorderIndexMap,
bool& notPossible)
{
if (inStart > inEnd)
return NULL;
// build the current Node
int rootData = preorder[preIndex];
Node* root = new Node(rootData);
preIndex++;
// find the node in inorderIndexMap
if (inorderIndexMap.find(rootData)
== inorderIndexMap.end()) {
notPossible = true;
return root;
}
int inorderIndex = inorderIndexMap[rootData];
if (!(inStart <= inorderIndex
&& inorderIndex <= inEnd)) {
notPossible = true;
return root;
}
int leftInorderStart = inStart,
leftInorderEnd = inorderIndex - 1,
rightInorderStart = inorderIndex + 1,
rightInorderEnd = inEnd;
root->left = buildTreeFromInorderPreorder(
leftInorderStart, leftInorderEnd, preIndex,
preorder, inorderIndexMap, notPossible);
if (notPossible)
return root;
root->right = buildTreeFromInorderPreorder(
rightInorderStart, rightInorderEnd, preIndex,
preorder, inorderIndexMap, notPossible);
return root;
}
bool checkPostorderCorrect(Node* root, int& postIndex,
int postorder[])
{
if (!root)
return true;
if (!checkPostorderCorrect(root->left, postIndex,
postorder))
return false;
if (!checkPostorderCorrect(root->right, postIndex,
postorder))
return false;
return (root->data == postorder[postIndex++]);
}
void printPostorder(Node* root)
{
if (!root)
return;
printPostorder(root->left);
printPostorder(root->right);
cout << root->data << ", ";
}
void printInorder(Node* root)
{
if (!root)
return;
printInorder(root->left);
cout << root->data << ", ";
printInorder(root->right);
}
bool checktree(int preorder[], int inorder[],
int postorder[], int N)
{
// Your code goes here
if (N == 0)
return true;
unordered_map<int, int> inorderIndexMap;
for (int i = 0; i < N; i++)
inorderIndexMap[inorder[i]] = i;
int preIndex = 0;
// return checkInorderPreorder(0, N - 1, preIndex,
// preorder, inorderIndexMap) &&
// checkInorderPostorder(0, N - 1, postIndex, postorder,
// inorderIndexMap);
bool notPossible = false;
Node* root = buildTreeFromInorderPreorder(
0, N - 1, preIndex, preorder, inorderIndexMap,
notPossible);
if (notPossible)
return false;
int postIndex = 0;
return checkPostorderCorrect(root, postIndex,
postorder);
}
// Driver program to test above functions
int main()
{
int inOrder[] = { 4, 2, 5, 1, 3 };
int preOrder[] = { 1, 2, 4, 5, 3 };
int postOrder[] = { 4, 5, 2, 3, 1 };
int len = sizeof(inOrder) / sizeof(inOrder[0]);
// If both postorder traversals are same
if (checktree(preOrder, inOrder, postOrder, len))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java code for the above approach
import java.util.*;
class Node {
int data;
Node left, right;
Node(int val)
{
data = val;
left = right = null;
}
}
class Main {
static Node buildTreeFromInorderPreorder(
int inStart, int inEnd, int[] preorder,
Map<Integer, Integer> inorderIndexMap,
boolean[] notPossible, int preIndex)
{
if (inStart > inEnd)
return null;
// build the current Node
int rootData = preorder[preIndex];
Node root = new Node(rootData);
preIndex++;
// find the node in inorderIndexMap
if (!inorderIndexMap.containsKey(rootData)) {
notPossible[0] = true;
return root;
}
int inorderIndex = inorderIndexMap.get(rootData);
if (!(inStart <= inorderIndex
&& inorderIndex <= inEnd)) {
notPossible[0] = true;
return root;
}
int leftInorderStart = inStart,
leftInorderEnd = inorderIndex - 1,
rightInorderStart = inorderIndex + 1,
rightInorderEnd = inEnd;
root.left = buildTreeFromInorderPreorder(
leftInorderStart, leftInorderEnd, preorder,
inorderIndexMap, notPossible, preIndex);
if (notPossible[0])
return root;
root.right = buildTreeFromInorderPreorder(
rightInorderStart, rightInorderEnd, preorder,
inorderIndexMap, notPossible, preIndex);
return root;
}
static boolean checkPostorderCorrect(Node root,
int[] postorder,
int postIndex)
{
if (root == null)
return true;
if (!checkPostorderCorrect(root.left, postorder,
postIndex))
return false;
if (!checkPostorderCorrect(root.right, postorder,
postIndex))
return false;
return (root.data == postorder[postIndex++]);
}
static boolean checktree(int[] preorder, int[] inorder,
int[] postorder, int N)
{
// Your code goes here
if (N == 0)
return true;
Map<Integer, Integer> inorderIndexMap
= new HashMap<>();
for (int i = 0; i < N; i++)
inorderIndexMap.put(inorder[i], i);
int preIndex = 0;
boolean[] notPossible = new boolean[] { false };
Node root = buildTreeFromInorderPreorder(
0, N - 1, preorder, inorderIndexMap,
notPossible, preIndex);
if (notPossible[0])
return true;
int postIndex = 0;
return checkPostorderCorrect(root, postorder,
postIndex);
}
public static void main(String[] args)
{
int inOrder[] = { 4, 2, 5, 1, 3 };
int preOrder[] = { 1, 2, 4, 5, 3 };
int postOrder[] = { 4, 5, 2, 3, 1 };
int len = inOrder.length;
if (checktree(preOrder, inOrder, postOrder, len))
System.out.println("The tree is valid");
else
System.out.println("The tree is not valid");
}
}
// This code is contributed by lokeshpotta20.
# Python3 program for the above approach
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
preIndex = 0
notPossible = False
def buildTreeFromInorderPreorder(inStart, inEnd, preorder, inorderIndexMap):
if(inStart > inEnd):
return None
# build the current node
global preIndex
global notPossible
rootData = preorder[preIndex]
root = Node(rootData)
preIndex += 1
# find the node in inorderIndexMap
if(inorderIndexMap.get(rootData) == None):
notPossible = True
return root
inorderIndex = inorderIndexMap.get(rootData)
if((inStart <= inorderIndex and inorderIndex <= inEnd) == False):
notPossible = True
return root
leftInorderStart = inStart
leftInorderEnd = inorderIndex - 1
rightInorderStart = inorderIndex + 1
rightInroderEnd = inEnd
root.left = buildTreeFromInorderPreorder(
leftInorderStart, leftInorderEnd, preorder, inorderIndexMap)
if(notPossible == True):
return root
root.right = buildTreeFromInorderPreorder(
rightInorderStart, rightInroderEnd, preorder, inorderIndexMap)
return root
postIndex = 0
def checkPostorderCorrect(root, postOrder):
if(root == None):
return True
if(checkPostorderCorrect(root.left, postOrder) == False):
return False
if(checkPostorderCorrect(root.right, postOrder) == False):
return False
global postIndex
if(root.data == postOrder[postIndex]):
postIndex += 1
return True
else:
postIndex += 1
return False
def printPostorder(root):
if(root == None):
return
printPostorder(root.left)
printPostorder(root.right)
print(root.data)
def printInorder(root):
if(root == None):
return
printInorder(root.left)
print(root.data)
printInorder(root.right)
def checktree(preorder, inorder, postorder, N):
if(N == 0):
return True
inorderIndexMap = {}
for i in range(N):
inorderIndexMap[inorder[i]] = i
root = buildTreeFromInorderPreorder(0, N-1, preorder, inorderIndexMap)
global notPossible
if(notPossible == True):
return False
if(checkPostorderCorrect(root, postorder)):
return True
else:
return False
# driver program
inOrder = [4, 2, 5, 1, 3]
preOrder = [1, 2, 4, 5, 3]
postOrder = [4, 5, 2, 3, 1]
len = len(inOrder)
# if both postorder traversal as same
if(checktree(preOrder, inOrder, postOrder, len) == True):
print("Yes")
else:
print("No")
# THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)
using System;
using System.Collections.Generic;
class Node
{
public int data;
public Node left;
public Node right;
public Node(int val)
{
data = val;
left = right = null;
}
}
class Program
{
static Node BuildTreeFromInorderPreorder(int inStart, int inEnd, int[] preorder, Dictionary<int, int> inorderIndexMap, ref bool notPossible, ref int preIndex)
{
if (inStart > inEnd)
return null;
// Build the current node
int rootData = preorder[preIndex];
Node root = new Node(rootData);
preIndex++;
// Find the node in inorderIndexMap
if (!inorderIndexMap.ContainsKey(rootData))
{
notPossible = true;
return root;
}
int inorderIndex = inorderIndexMap[rootData];
if (!(inStart <= inorderIndex && inorderIndex <= inEnd))
{
notPossible = true;
return root;
}
int leftInorderStart = inStart;
int leftInorderEnd = inorderIndex - 1;
int rightInorderStart = inorderIndex + 1;
int rightInorderEnd = inEnd;
root.left = BuildTreeFromInorderPreorder(leftInorderStart, leftInorderEnd, preorder, inorderIndexMap, ref notPossible, ref preIndex);
if (notPossible)
return root;
root.right = BuildTreeFromInorderPreorder(rightInorderStart, rightInorderEnd, preorder, inorderIndexMap, ref notPossible, ref preIndex);
return root;
}
static bool CheckPostorderCorrect(Node root, int[] postorder, ref int postIndex)
{
if (root == null)
return true;
if (!CheckPostorderCorrect(root.left, postorder, ref postIndex))
return false;
if (!CheckPostorderCorrect(root.right, postorder, ref postIndex))
return false;
return (root.data == postorder[postIndex++]);
}
static bool CheckTree(int[] preorder, int[] inorder, int[] postorder, int N)
{
if (N == 0)
return true;
Dictionary<int, int> inorderIndexMap = new Dictionary<int, int>();
for (int i = 0; i < N; i++)
inorderIndexMap.Add(inorder[i], i);
int preIndex = 0;
bool notPossible = false;
Node root = BuildTreeFromInorderPreorder(0, N - 1, preorder, inorderIndexMap, ref notPossible, ref preIndex);
if (notPossible)
return false;
int postIndex = 0;
return CheckPostorderCorrect(root, postorder, ref postIndex);
}
static void Main(string[] args)
{
int[] inOrder = { 4, 2, 5, 1, 3 };
int[] preOrder = { 1, 2, 4, 5, 3 };
int[] postOrder = { 4, 5, 2, 3, 1 };
int len = inOrder.Length;
if (CheckTree(preOrder, inOrder, postOrder, len))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// JavaScript program for the above approach
class Node{
constructor(val){
this.data = val;
this.left = null;
this.right = null;
}
}
let preIndex = 0;
let notPossible = false;
function buildTreeFromInorderPreorder(inStart, inEnd,
preorder, inorderIndexMap){
if(inStart > inEnd){
return null;
}
// build the current node
let rootData = preorder[preIndex];
let root = new Node(rootData);
preIndex++;
// find the node in inorderIndexMap
if(inorderIndexMap.has(rootData) == false){
notPossible = true;
return root;
}
let inorderIndex = inorderIndexMap.get(rootData);
if(!(inStart <= inorderIndex && inorderIndex <= inEnd)){
notPosstible = true;
return root;
}
let leftInorderStart = inStart;
let leftInorderEnd = inorderIndex - 1;
let rightInorderStart = inorderIndex + 1;
let rightInorderEnd = inEnd;
root.left = buildTreeFromInorderPreorder(
leftInorderStart, leftInorderEnd, preorder, inorderIndexMap);
if(notPossible == true)
return root;
root.right = buildTreeFromInorderPreorder(
leftInorderStart, leftInorderEnd, preorder, inorderIndexMap);
return root;
}
let postIndex = 0;
function checkPostorderCorrect(root, postOrder){
if(root == null) return true;
if(!checkPostorderCorrect(root.left, postOrder))
return false;
if(!checkPostorderCorrect(root.right, postOrder))
return false;
return (root.data == postOrder[postIndex++]);
}
function printPostorder(root){
if(root == null) return;
printPostorder(root.left);
printPostorder(root.right);
document.write(root.data + " ");
}
function printInorder(root){
if(root == null) return;
printInorder(root.left);
document.write(root.data + " ");
printInorder(root.right);
}
function checktree(preorder, inorder, postorder, N){
if(N == 0) return true;
inorderIndexMap = new Map();
for(let i = 0; i<N; i++){
inorderIndexMap.set(inorder[i], i);
}
let root = buildTreeFromInorderPreorder(0, N-1,
preorder, inorderIndexMap);
if(notPossible == true) return false;
return checkPostorderCorrect(root, postorder);
}
// driver program
let inOrder = [4, 2, 5, 1, 3];
let preOrder = [1, 2, 4, 5, 3];
let postOrder = [4, 5, 2, 3, 1];
let len = inOrder.length;
// If both postorder traversals as same
if(checktree(preOrder, inOrder, postOrder, len))
document.write("Yes");
else
document.write("No");
// This code is contributed by Yash Agarwal(yashagarwal2852002)
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(N), where N is number of nodes in the tree.
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