Open in App
Not now

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

• Difficulty Level : Medium
• Last Updated : 31 Jan, 2023

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, the 2nd ancestor of 5 is 1. 3rd ancestor of node 5 will be -1.

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 the kth ancestor. Once we have reached the kth parent, we will simply print the node and return NULL.

Below is the implementation of the above idea:

## C++

 `/* C++ program to calculate Kth ancestor of given node */` `#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;` `}`

## Java

 `// Java program to calculate Kth ancestor of given node ` `class` `Solution` `{`   `// A Binary Tree Node` `static` `class` `Node` `{` `    ``int` `data;` `    ``Node left, right;` `};`   `// temporary node to keep track of Node returned` `// from previous recursive call during backtrack` `static` `Node temp = ``null``;` `static` `int` `k;`   `// recursive function to calculate Kth ancestor` `static` `Node kthAncestorDFS(Node root, ``int` `node )` `{ ` `    ``// Base case` `    ``if` `(root == ``null``)` `        ``return` `null``;` `    `  `    ``if` `(root.data == node||` `    ``(temp = kthAncestorDFS(root.left,node)) != ``null` `||` `    ``(temp = kthAncestorDFS(root.right,node)) != ``null``)` `    ``{ ` `        ``if` `(k > ``0``)     ` `            ``k--;` `        `  `        ``else` `if` `(k == ``0``)` `        ``{` `            ``// print the kth ancestor` `            ``System.out.print(``"Kth ancestor is: "``+root.data);` `            `  `            ``// return null to stop further backtracking` `            ``return` `null``;` `        ``}` `        `  `        ``// return current node to previous call` `        ``return` `root;` `    ``}` `    ``return` `null``;` `} `   `// Utility function to create a new tree node` `static` `Node newNode(``int` `data)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.data = data;` `    ``temp.left = temp.right = ``null``;` `    ``return` `temp;` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``// 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``);`   `    ``k = ``2``;` `    ``int` `node = ``5``;`   `    ``// print kth ancestor of given node` `    ``Node parent = kthAncestorDFS(root,node);` `    `  `    ``// check if parent is not null, it means` `    ``// there is no Kth ancestor of the node` `    ``if` `(parent != ``null``)` `        ``System.out.println(``"-1"``);` `}` `}`   `// This code is contributed by Arnab Kundu`

## 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` `        `  `        ``else` `if` `(k[``0``] ``=``=` `0``):` `            `  `            ``# print the 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`   `    ``# print kth 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`

## C#

 `// C# program to calculate Kth ancestor of given node ` `using` `System;` `    `  `class` `GFG` `{`   `// A Binary Tree Node` `public` `class` `Node` `{` `    ``public` `int` `data;` `    ``public` `Node left, right;` `};`   `// temporary node to keep track of Node returned` `// from previous recursive call during backtrack` `static` `Node temp = ``null``;` `static` `int` `k;`   `// recursive function to calculate Kth ancestor` `static` `Node kthAncestorDFS(Node root, ``int` `node )` `{ ` `    ``// Base case` `    ``if` `(root == ``null``)` `        ``return` `null``;` `    `  `    ``if` `(root.data == node||` `    ``(temp = kthAncestorDFS(root.left,node)) != ``null` `||` `    ``(temp = kthAncestorDFS(root.right,node)) != ``null``)` `    ``{ ` `        ``if` `(k > 0)     ` `            ``k--;` `        `  `        ``else` `if` `(k == 0)` `        ``{` `            ``// print the kth ancestor` `            ``Console.Write(``"Kth ancestor is: "``+root.data);` `            `  `            ``// return null to stop further backtracking` `            ``return` `null``;` `        ``}` `        `  `        ``// return current node to previous call` `        ``return` `root;` `    ``}` `    ``return` `null``;` `} `   `// Utility function to create a new tree node` `static` `Node newNode(``int` `data)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.data = data;` `    ``temp.left = temp.right = ``null``;` `    ``return` `temp;` `}`   `// Driver code` `public` `static` `void` `Main(String []args)` `{` `    ``// 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);`   `    ``k = 2;` `    ``int` `node = 5;`   `    ``// print kth ancestor of given node` `    ``Node parent = kthAncestorDFS(root,node);` `    `  `    ``// check if parent is not null, it means` `    ``// there is no Kth ancestor of the node` `    ``if` `(parent != ``null``)` `        ``Console.WriteLine(``"-1"``);` `}` `}`   `// This code is contributed by 29AjayKumar `

## Javascript

 ``

Output

`Kth ancestor is: 1`

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

Auxiliary Space: O(h) where h is the height of binary tree due to recursion call.

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up
Related Articles