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Ancestors in Binary Tree

Ancestors in Binary Tree: In this article, we are going to see how to find ancestors for a node in binary tree?
Submitted by Radib Kar, on March 25, 2019

Problem statement:

Given a Binary Tree and a target key, write a function that prints all the ancestors of the key in the given binary tree.

Example:

Let's the tree be like following:

Ancestors in Binary Tree
    Let for node value 12:
    Ancestors are:
    7, 5, 8
    While for node value 7:
    Ancestors are:
    5, 8

Solution

What is Ancestors?

For any node n,
Its ancestors are the nodes which are on the path between roots to node n

Thus for the above examples,

Example 1:

    Node is 12 //represented by value
    Root to the node path is
    8->5->7->12
    Thus the ancestors are 7, 5, 8

Example 2:

    Node is 7 //represented by value
    Root to the node path is
    8->5->7
    Thus the ancestors are 5, 8

Algorithm:

FUNCTION printAncestors(Node *root, int target)
    IF(!root)
        return false;
    IF( (root->left && root->left->data==target) ||
        (root->right && root->right->data==target ) || 
        printAncestors(root->left,target)|| 
        printAncestors(root->right,target)){
            Print root->data;
            return true;
    END IF
    return false;
END FUNCTION

That simply means we are doing kind of DFS

For a currentnode to be ancestor of the target node the conditions are:

1.  If the target node is its child node (either left child or right child)
    //condition
    IF((root->left && root->left->data==target) ||
    (root->right && root->right->data==target ))

2.  If any of the two subtree of the current node contain ancestor of the target 
    node then the current node is also an ancestor. 
    //condition
    IF(printAncestors(root->left, target)|| 
    printAncestors(root->right, target))

Example with explanation:

For target node 7:
Root 8 is ancestor on condition: its left subtree contains ancestor 5
5 is ancestor since target node is its right child
Thus ancestors are:
5, 8 //in order

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

//tree node
class Node{ 
	public:
	int data;
	Node *left;
	Node *right;
};

bool printAncestors(Node *root, int target)
{
	if(!root)
	return false;

	if(	(root->left && root->left->data==target) ||
		(root->right && root->right->data==target ) || 
		printAncestors(root->left,target)|| 
		printAncestors(root->right,target)){
		cout<<root->data<<" ";
		return true;
	}
	return false;
}

//creating new nodes
Node* newnode(int data){  
	Node* node = (Node*)malloc(sizeof(Node)); 
	node->data = data; 
	node->left = NULL; 
	node->right = NULL; 
	return(node); 
} 

int main() { 
	//**same tree is builted as shown in example**
	cout<<"tree in the example is build here"<<endl;
	//building the tree like as in the example
	Node *root=newnode(8); 
	root->left= newnode(5); 
	root->right= newnode(4); 
	root->right->right=newnode(11);
	root->right->right->left=newnode(3);
	root->left->left=newnode(9); 
	root->left->right=newnode(7);
	root->left->right->left=newnode(1);
	root->left->right->right=newnode(12);
	root->left->right->right->left=newnode(2);

	int s;

	cout<<"enter input value to find ancestors......"<<endl;
	cin>>s;

	printAncestors(root,s);
	return 0; 
}

Output

tree in the example is build here
enter input value to find ancestors......
7
5 8





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