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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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 corresponding row in 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 = 100mat = [[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 |
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
Time complexity of above solution is O(n2).
How to do reverse – construct tree from ancestor matrix?
Construct tree from ancestor matrix
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