Construct Ancestor Matrix from a Given Binary Tree
Given a Binary Tree where all values are from 0 to n-1. Construct an ancestor matrix mat[n][n]. Ancestor matrix is defined as below.
mat[i][j] = 1 if i is ancestor of j mat[i][j] = 0, otherwise
Examples:
Input: Root of below Binary Tree. 0 / \ 1 2 Output: 0 1 1 0 0 0 0 0 0 Input: Root of below Binary Tree. 5 / \ 1 2 / \ / 0 4 3 Output: 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0
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Method 1: The idea is to traverse the tree. While traversing, keep track of ancestors in an array. When we visit a node, we add it to ancestor array and consider the corresponding row in the adjacency matrix. We mark all ancestors in its row as 1. Once a node and all its children are processed, we remove the node from ancestor array.
Below is the implementation of above idea.
C++
// C++ program to construct ancestor matrix for // given tree. #include<bits/stdc++.h> using namespace std; #define MAX 100 /* A binary tree node */ struct Node { int data; Node *left, *right; }; // Creating a global boolean matrix for simplicity bool mat[MAX][MAX]; // anc[] stores all ancestors of current node. This // function fills ancestors for all nodes. // It also returns size of tree. Size of tree is // used to print ancestor matrix. int ancestorMatrixRec(Node *root, vector< int > &anc) { /* base case */ if (root == NULL) return 0;; // Update all ancestors of current node int data = root->data; for ( int i=0; i<anc.size(); i++) mat[anc[i]][data] = true ; // Push data to list of ancestors anc.push_back(data); // Traverse left and right subtrees int l = ancestorMatrixRec(root->left, anc); int r = ancestorMatrixRec(root->right, anc); // Remove data from list the list of ancestors // as all descendants of it are processed now. anc.pop_back(); return l+r+1; } // This function mainly calls ancestorMatrixRec() void ancestorMatrix(Node *root) { // Create an empty ancestor array vector< int > anc; // Fill ancestor matrix and find size of // tree. int n = ancestorMatrixRec(root, anc); // Print the filled values for ( int i=0; i<n; i++) { for ( int j=0; j<n; j++) cout << mat[i][j] << " " ; cout << endl; } } /* Helper function to create a new node */ Node* newnode( int data) { Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } /* Driver program to test above functions*/ int main() { /* Construct the following binary tree 5 / \ 1 2 / \ / 0 4 3 */ Node *root = newnode(5); root->left = newnode(1); root->right = newnode(2); root->left->left = newnode(0); root->left->right = newnode(4); root->right->left = newnode(3); ancestorMatrix(root); return 0; } |
Java
// Java Program to construct ancestor matrix for a given tree import java.util.*; class GFG { // ancestorMatrix function to populate the matrix of public static void ancestorMatrix(Node root , int matrix[][], int size) { // base case: if (root== null ) return ; // call recursively for a preorder {left} ancestorMatrix(root.left, matrix, size); // call recursively for preorder {right} ancestorMatrix(root.right, matrix, size); // here we will reach the root node automatically // try solving on pen and paper if (root.left != null ) { // make the current node as parent of its children node matrix[root.data][root.left.data] = 1 ; // iterate through all the columns of children node // all nodes which are children to // children of root node will also // be children of root node for ( int i = 0 ; i < size; i++) { // if children of root node is a parent // of someone (i.e 1) then make that node // as children of root also if (matrix[root.left.data][i] == 1 ) matrix[root.data][i] = 1 ; } } // same procedure followed for right node as well if (root.right != null ) { matrix[root.data][root.right.data] = 1 ; for ( int i = 0 ; i < size; i++) { if (matrix[root.right.data][i]== 1 ) matrix[root.data][i] = 1 ; } } } // Driver program to test the program public static void main(String[] args) { // construct the binary tree as follows Node tree_root = new Node( 5 ); tree_root.left = new Node ( 1 ); tree_root.right = new Node( 2 ); tree_root.left.left = new Node( 0 ); tree_root.left.right = new Node( 4 ); tree_root.right.left = new Node( 3 ); // size of matrix int size = 6 ; int matrix [][] = new int [size][size]; ancestorMatrix(tree_root, matrix, size); for ( int i = 0 ; i < size; i++) { for ( int j = 0 ; j < size; j++) { System.out.print(matrix[i][j]+ " " ); } System.out.println(); } } // node class for tree node static class Node { public int data ; public Node left ,right; public Node ( int data) { this .data = data; this .left = this .right = null ; } } } // This code is contributed by Sparsh Singhal |
Python3
# Python3 program to construct ancestor # matrix for given tree. class newnode: def __init__( self , data): self .data = data self .left = self .right = None # anc[] stores all ancestors of current node. # This function fills ancestors for all nodes. # It also returns size of tree. Size of tree # is used to print ancestor matrix. def ancestorMatrixRec(root, anc): global mat, MAX # base case if root = = None : return 0 # Update all ancestors of current node data = root.data for i in range ( len (anc)): mat[anc[i]][data] = 1 # Push data to list of ancestors anc.append(data) # Traverse left and right subtrees l = ancestorMatrixRec(root.left, anc) r = ancestorMatrixRec(root.right, anc) # Remove data from list the list of ancestors # as all descendants of it are processed now. anc.pop( - 1 ) return l + r + 1 # This function mainly calls ancestorMatrixRec() def ancestorMatrix(root): # Create an empty ancestor array anc = [] # Fill ancestor matrix and find # size of tree. n = ancestorMatrixRec(root, anc) # Print the filled values for i in range (n): for j in range (n): print (mat[i][j], end = " " ) print () # Driver Code MAX = 100 mat = [[ 0 ] * MAX for i in range ( MAX )] # Construct the following binary tree # 5 # / \ # 1 2 # / \ / # 0 4 3 root = newnode( 5 ) root.left = newnode( 1 ) root.right = newnode( 2 ) root.left.left = newnode( 0 ) root.left.right = newnode( 4 ) root.right.left = newnode( 3 ) ancestorMatrix(root) # This code is contributed by PranchalK |
C#
// C# Program to construct ancestor matrix for a given tree using System; class GFG { // ancestorMatrix function to populate the matrix of public static void ancestorMatrix(Node root, int [,]matrix, int size) { // base case: if (root == null ) { return ; } // call recursively for a preorder {left} ancestorMatrix(root.left, matrix, size); // call recursively for preorder {right} ancestorMatrix(root.right, matrix, size); // here we will reach the root node automatically // try solving on pen and paper if (root.left != null ) { // make the current node as parent of its children node matrix[root.data,root.left.data] = 1; // iterate through all the columns of children node // all nodes which are children to // children of root node will also // be children of root node for ( int i = 0; i < size; i++) { // if children of root node is a parent // of someone (i.e 1) then make that node // as children of root also if (matrix[root.left.data,i] == 1) { matrix[root.data,i] = 1; } } } // same procedure followed for right node as well if (root.right != null ) { matrix[root.data,root.right.data] = 1; for ( int i = 0; i < size; i++) { if (matrix[root.right.data,i] == 1) { matrix[root.data,i] = 1; } } } } // Driver code public static void Main(String[] args) { // construct the binary tree as follows Node tree_root = new Node(5); tree_root.left = new Node(1); tree_root.right = new Node(2); tree_root.left.left = new Node(0); tree_root.left.right = new Node(4); tree_root.right.left = new Node(3); // size of matrix int size = 6; int [,]matrix = new int [size,size]; ancestorMatrix(tree_root, matrix, size); for ( int i = 0; i < size; i++) { for ( int j = 0; j < size; j++) { Console.Write(matrix[i,j] + " " ); } Console.WriteLine(); } } // node class for tree node public class Node { public int data; public Node left, right; public Node( int data) { this .data = data; this .left = this .right = null ; } } } // This code is contributed by 29AjayKumar |
Javascript
<script> class Node { constructor(data) { this .data=data; this .left = this .right = null ; } } function ancestorMatrixRec(root ,anc) { // base case: if (root== null ) return 0; let data = root.data; for (let i=0;i<anc.length;i++) { matrix[anc[i]][data] = 1; } //document.write(data+"<br>") // Push data to list of ancestors anc.push(data); // Traverse left and right subtrees let l = ancestorMatrixRec(root.left, anc); let r = ancestorMatrixRec(root.right, anc); // Remove data from list the list of ancestors // as all descendants of it are processed now. anc.pop() return l + r + 1; } function ancestorMatrix(root) { let anc = []; let n = ancestorMatrixRec(root, anc); for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { document.write(matrix[i][j]+ " " ); } document.write( "<br>" ); } } // construct the binary tree as follows let tree_root = new Node(5); tree_root.left = new Node (1); tree_root.right = new Node(2); tree_root.left.left = new Node(0); tree_root.left.right = new Node(4); tree_root.right.left = new Node(3); // size of matrix let size = 100; let matrix = new Array(size); for (let i=0;i<size;i++) { matrix[i]= new Array(size); for (let j=0;j<size;j++) { matrix[i][j]=0; } } ancestorMatrix(tree_root); // This code is contributed by rag2127 </script> |
0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0
Time complexity of above solution is O(n2).
Auxiliary Space: O(n2), since n2 extra space has been taken.
Method 2:
This method doesn’t use any auxiliary space to store values in the vector. Create a 2D matrix( say M) of the given size. Now the idea is to traverse the tree in PreOrder. While traversing, keep track of the last ancestor.
When we visit a node, if the node is NULL return, else assign M[lastAncestorValue][currentNodeValue]=1.
Through this, we have the Matrix(M) which keeps tracks of the Lowest Ancestor, Now by applying transitive closure to this Matrix(M), we get the required result.
Below is the implementation of the above idea.
C++
// C++ program to construct ancestor matrix for // given tree. #include<bits/stdc++.h> using namespace std; #define size 6 int M[size][size]={0}; /* A binary tree node */ struct Node { int data; Node *left, *right; }; /* Helper function to create a new node */ Node* newnode( int data) { Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } void printMatrix(){ for ( int i=0;i<size;i++){ for ( int j=0;j<size;j++) cout<<M[i][j]<< " " ; cout<<endl; } } //First PreOrder Traversal void MatrixUtil(Node *root, int index){ if (root==NULL) return ; int preData=root->data; //Since there is no ancestor for root node, // so we doesn't assign it's value as 1 if (index==-1)index=root->data; else M[index][preData]=1; MatrixUtil(root->left,preData); MatrixUtil(root->right,preData); } void Matrix(Node *root){ // Call Func MatrixUtil MatrixUtil(root,-1); //Applying Transitive Closure for the given Matrix for ( int i=0;i<size;i++){ for ( int j = 0;j< size; j++) { for ( int k=0;k<size;k++) M[j][k]=M[j][k]||(M[j][i]&&M[i][k]); } } //Printing Matrix printMatrix(); } /* Driver program to test above functions*/ int main() { Node *root = newnode(5); root->left = newnode(1); root->right = newnode(2); root->left->left = newnode(0); root->left->right = newnode(4); root->right->left = newnode(3); Matrix(root); return 0; } |
Java
// Java program to conancestor matrix for // given tree. import java.util.*; class GFG { static final int size = 6 ; static int [][]M = new int [size][size]; /* A binary tree node */ static class Node { int data; Node left, right; }; /* Helper function to create a new node */ static Node newnode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } static void printMatrix() { for ( int i = 0 ; i < size; i++) { for ( int j = 0 ; j < size; j++) System.out.print(M[i][j] + " " ); System.out.println(); } } // First PreOrder Traversal static void MatrixUtil(Node root, int index) { if (root == null ) return ; int preData = root.data; // Since there is no ancestor for root node, // so we doesn't assign it's value as 1 if (index == - 1 )index = root.data; else M[index][preData] = 1 ; MatrixUtil(root.left, preData); MatrixUtil(root.right, preData); } static void Matrix(Node root) { // Call Func MatrixUtil MatrixUtil(root, - 1 ); // Applying Transitive Closure for the given Matrix for ( int i = 0 ; i < size; i++) { for ( int j = 0 ; j < size; j++) { for ( int k = 0 ; k < size; k++) { if (M[j][k] == 1 ) M[j][k] = 1 ; else if ((M[j][i] == 1 && M[i][k] == 1 )) M[j][k] = 1 ; } } } // Printing Matrix printMatrix(); } /* Driver program to test above functions*/ public static void main(String[] args) { Node root = newnode( 5 ); root.left = newnode( 1 ); root.right = newnode( 2 ); root.left.left = newnode( 0 ); root.left.right = newnode( 4 ); root.right.left = newnode( 3 ); Matrix(root); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 program to construct ancestor # matrix for given tree. size = 6 M = [[ 0 for j in range (size)] for i in range (size)] # A binary tree node class Node: def __init__( self , data): self .left = None self .right = None self .data = data # Helper function to create a new node def newnode(data): temp = Node(data) return temp def printMatrix(): for i in range (size): for j in range (size): print (M[i][j], end = ' ' ) print () # First PreOrder Traversal def MatrixUtil(root, index): if (root = = None ): return preData = root.data # Since there is no ancestor for # root node, so we doesn't assign # it's value as 1 if (index = = - 1 ): index = root.data else : M[index][preData] = 1 MatrixUtil(root.left, preData) MatrixUtil(root.right, preData) def Matrix(root): # Call Func MatrixUtil MatrixUtil(root, - 1 ) # Applying Transitive Closure # for the given Matrix for i in range (size): for j in range (size): for k in range (size): M[j][k] = (M[j][k] or (M[j][i] and M[i][k])) # Printing Matrix printMatrix() # Driver code if __name__ = = "__main__" : root = newnode( 5 ) root.left = newnode( 1 ) root.right = newnode( 2 ) root.left.left = newnode( 0 ) root.left.right = newnode( 4 ) root.right.left = newnode( 3 ) Matrix(root) # This code is contributed by rutvik_56 |
C#
// C# program to conancestor matrix for // given tree. using System; class GFG { static readonly int size = 6; static int [,]M = new int [size, size]; /* A binary tree node */ public class Node { public int data; public Node left, right; }; /* Helper function to create a new node */ static Node newnode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } static void printMatrix() { for ( int i = 0; i < size; i++) { for ( int j = 0; j < size; j++) Console.Write(M[i, j] + " " ); Console.WriteLine(); } } // First PreOrder Traversal static void MatrixUtil(Node root, int index) { if (root == null ) return ; int preData = root.data; // Since there is no ancestor for root node, // so we doesn't assign it's value as 1 if (index == -1)index = root.data; else M[index,preData] = 1; MatrixUtil(root.left, preData); MatrixUtil(root.right, preData); } static void Matrix(Node root) { // Call Func MatrixUtil MatrixUtil(root, -1); // Applying Transitive Closure for the given Matrix for ( int i = 0; i < size; i++) { for ( int j = 0; j < size; j++) { for ( int k = 0; k < size; k++) { if (M[j, k] == 1) M[j, k] = 1; else if ((M[j, i] == 1 && M[i, k] == 1)) M[j, k] = 1; } } } // Printing Matrix printMatrix(); } /* Driver code*/ public static void Main(String[] args) { Node root = newnode(5); root.left = newnode(1); root.right = newnode(2); root.left.left = newnode(0); root.left.right = newnode(4); root.right.left = newnode(3); Matrix(root); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript program to conancestor matrix for // given tree. var size = 6; var M = Array(size).fill().map(()=>Array(size).fill(0)); /* A binary tree node */ class Node { constructor(val) { this .data = val; this .left = null ; this .right = null ; } } /* Helper function to create a new node */ function newnode(data) { var node = new Node(); node.data = data; node.left = node.right = null ; return (node); } function printMatrix() { for (i = 0; i < size; i++) { for (j = 0; j < size; j++) document.write(M[i][j] + " " ); document.write( "<br/>" ); } } // First PreOrder Traversal function MatrixUtil(root , index) { if (root == null ) return ; var preData = root.data; // Since there is no ancestor for root node, // so we doesn't assign it's value as 1 if (index == -1) index = root.data; else M[index][preData] = 1; MatrixUtil(root.left, preData); MatrixUtil(root.right, preData); } function Matrix(root) { // Call Func MatrixUtil MatrixUtil(root, -1); // Applying Transitive Closure for the given Matrix for (i = 0; i < size; i++) { for (j = 0; j < size; j++) { for (k = 0; k < size; k++) { if (M[j][k] == 1) M[j][k] = 1; else if ((M[j][i] == 1 && M[i][k] == 1)) M[j][k] = 1; } } } // Printing Matrix printMatrix(); } /* Driver program to test above functions */ var root = newnode(5); root.left = newnode(1); root.right = newnode(2); root.left.left = newnode(0); root.left.right = newnode(4); root.right.left = newnode(3); Matrix(root); // This code is contributed by gauravrajput1 </script> |
0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0
Time complexity of above solution is O(n2).
Auxiliary Space: O(n2)
How to do reverse – construct tree from ancestor matrix?
Construct tree from ancestor matrix
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