Construct Ancestor Matrix from a Given Binary Tree

Last Updated : 27 Aug, 2022

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

We strongly recommend you to minimize your browser and try this yourself first.

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>

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).
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>

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).
Auxiliary Space: O(n²)

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