# Kth ancestor of a node in binary tree | Set 2

Given a binary tree in which nodes are numbered from 1 to n. Given a node and a positive integer K. We have to print the Kth ancestor of the given node in the binary tree. If there does not exist any such ancestor then print -1.

For example in the below given binary tree, 2nd ancestor of node 4 and 5 is 1. 3rd ancestor of node 4 will be -1. ## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed a BFS based solution for this problem in our previous article. If you observe that solution carefully, you will see that the basic approach was to first find the node and then backtrack to the kth parent. The same thing can be done using recursive DFS without using an extra array.
The idea of using DFS is to first find the given node in the tree, and then backtrack k times to reach to kth ancestor, once we have reached to the kth parent, we will simply print the node and return NULL.

Below is the implementation of above idea:

## C++

 `/* C++ program to calculate Kth ancestor of given node */` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// A Binary Tree Node ` `struct` `Node ` `{ ` `    ``int` `data; ` `    ``struct` `Node *left, *right; ` `}; ` ` `  `// temporary node to keep track of Node returned ` `// from previous recursive call during backtrack ` `Node* temp = NULL; ` ` `  `// recursive function to calculate Kth ancestor ` `Node* kthAncestorDFS(Node *root, ``int` `node , ``int` `&k) ` `{    ` `    ``// Base case ` `    ``if` `(!root) ` `        ``return` `NULL; ` `     `  `    ``if` `(root->data == node|| ` `       ``(temp =  kthAncestorDFS(root->left,node,k)) || ` `       ``(temp =  kthAncestorDFS(root->right,node,k))) ` `    ``{    ` `        ``if` `(k > 0)         ` `            ``k--; ` `         `  `        ``else` `if` `(k == 0) ` `        ``{ ` `            ``// print the kth ancestor ` `            ``cout<<``"Kth ancestor is: "``<data; ` `             `  `            ``// return NULL to stop further backtracking ` `            ``return` `NULL; ` `        ``} ` `         `  `        ``// return current node to previous call ` `        ``return` `root; ` `    ``} ` `}  ` ` `  `// 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; ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``// Let us create binary tree shown in above diagram ` `    ``Node *root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` ` `  `    ``int` `k = 2; ` `    ``int` `node = 5; ` ` `  `    ``// print kth ancestor of given node ` `    ``Node* parent = kthAncestorDFS(root,node,k); ` `     `  `    ``// check if parent is not NULL, it means ` `    ``// there is no Kth ancestor of the node ` `    ``if` `(parent) ` `        ``cout << ``"-1"``; ` `     `  `    ``return` `0; ` `} `

## Python3

 `""" Python3 program to calculate Kth  ` `    ``ancestor of given node """` ` `  `# A Binary Tree Node  ` `# Utility function to create a new tree node  ` `class` `newNode:  ` ` `  `    ``# Constructor to create a new node  ` `    ``def` `__init__(``self``, data):  ` `        ``self``.data ``=` `data  ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `# recursive function to calculate  ` `# Kth ancestor  ` `def` `kthAncestorDFS(root, node, k):  ` `     `  `    ``# Base case  ` `    ``if` `(``not` `root): ` `        ``return` `None` `     `  `    ``if` `(root.data ``=``=` `node ``or` `       ``(kthAncestorDFS(root.left, node, k)) ``or` `       ``(kthAncestorDFS(root.right, node, k))): ` `         `  `        ``if` `(k[``0``] > ``0``): ` `            ``k[``0``] ``-``=` `1` `         `  `        ``elif` `(k[``0``] ``=``=` `0``): ` `             `  `            ``# prthe kth ancestor  ` `            ``print``(``"Kth ancestor is:"``, root.data) ` `             `  `            ``# return None to stop further ` `            ``# backtracking  ` `            ``return` `None` `             `  `        ``# return current node to previous call  ` `        ``return` `root ` `     `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``root ``=` `newNode(``1``)  ` `    ``root.left ``=` `newNode(``2``)  ` `    ``root.right ``=` `newNode(``3``)  ` `    ``root.left.left ``=` `newNode(``4``)  ` `    ``root.left.right ``=` `newNode(``5``)  ` ` `  `    ``k ``=` `[``2``] ` `    ``node ``=` `5` ` `  `    ``# prkth ancestor of given node  ` `    ``parent ``=` `kthAncestorDFS(root,node,k)  ` `     `  `    ``# check if parent is not None, it means  ` `    ``# there is no Kth ancestor of the node  ` `    ``if` `(parent): ` `        ``print``(``"-1"``) ` ` `  `# This code is contributed  ` `# by SHUBHAMSINGH10 `

Output:

```Kth ancestor is: 1
```

Time Complexity : O(n), where n is the number of nodes in the binary tree.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

## tags:

Tree Binary Tree Tree

code

load comments