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# K-th ancestor of a node in Binary Tree

• Difficulty Level : Medium
• Last Updated : 20 Oct, 2022

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 K-th 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 Practice

The idea to do this is to first traverse the binary tree and store the ancestor of each node in an array of size n. For example, suppose the array is ancestor[n]. Then at index i, ancestor[i] will store the ancestor of ith node. So, the 2nd ancestor of ith node will be ancestor[ancestor[i]] and so on. We will use this idea to calculate the kth ancestor of the given node. We can use level order traversal to populate this array of ancestors.

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;` `};`   `// function to generate array of ancestors` `void` `generateArray(Node *root, ``int` `ancestors[])` `{` `    ``// There will be no ancestor of root node` `    ``ancestors[root->data] = -1;`   `    ``// level order traversal to ` `    ``// generate 1st ancestor` `    ``queue q;` `    ``q.push(root);`   `    ``while``(!q.empty())` `    ``{` `        ``Node* temp  = q.front();` `        ``q.pop();`   `        ``if` `(temp->left)` `        ``{` `            ``ancestors[temp->left->data] = temp->data;` `            ``q.push(temp->left);` `        ``}`   `        ``if` `(temp->right)` `        ``{` `            ``ancestors[temp->right->data] = temp->data;` `            ``q.push(temp->right);` `        ``}` `    ``} ` `}`   `// function to calculate Kth ancestor` `int` `kthAncestor(Node *root, ``int` `n, ``int` `k, ``int` `node)` `{` `    ``// create array to store 1st ancestors` `    ``int` `ancestors[n+1] = {0};`   `    ``// generate first ancestor array` `    ``generateArray(root,ancestors);`   `    ``// variable to track record of number of` `    ``// ancestors visited` `    ``int` `count = 0;`   `    ``while` `(node!=-1)` `    ``{   ` `        ``node = ancestors[node];` `        ``count++;`   `        ``if``(count==k)` `            ``break``;` `    ``}`   `    ``// print Kth ancestor` `    ``return` `node;` `} `   `// 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` `    ``cout<

## Java

 `/* Java program to calculate Kth ancestor of given node */` `import` `java.util.*; ` `class` `GfG {` `// A Binary Tree Node ` `static` `class` `Node ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` `}`   `// function to generate array of ancestors ` `static` `void` `generateArray(Node root, ``int` `ancestors[]) ` `{ ` `    ``// There will be no ancestor of root node ` `    ``ancestors[root.data] = -``1``; `   `    ``// level order traversal to ` `    ``// generate 1st ancestor ` `    ``Queue q = ``new` `LinkedList (); ` `    ``q.add(root); `   `    ``while``(!q.isEmpty()) ` `    ``{ ` `        ``Node temp = q.peek(); ` `        ``q.remove(); `   `        ``if` `(temp.left != ``null``) ` `        ``{ ` `            ``ancestors[temp.left.data] = temp.data; ` `            ``q.add(temp.left); ` `        ``} `   `        ``if` `(temp.right != ``null``) ` `        ``{ ` `            ``ancestors[temp.right.data] = temp.data; ` `            ``q.add(temp.right); ` `        ``} ` `    ``} ` `} `   `// function to calculate Kth ancestor ` `static` `int` `kthAncestor(Node root, ``int` `n, ``int` `k, ``int` `node) ` `{ ` `    ``// create array to store 1st ancestors ` `    ``int` `ancestors[] = ``new` `int``[n + ``1``]; `   `    ``// generate first ancestor array ` `    ``generateArray(root,ancestors); `   `    ``// variable to track record of number of ` `    ``// ancestors visited ` `    ``int` `count = ``0``; `   `    ``while` `(node!=-``1``) ` `    ``{ ` `        ``node = ancestors[node]; ` `        ``count++; `   `        ``if``(count==k) ` `            ``break``; ` `    ``} `   `    ``// print Kth ancestor ` `    ``return` `node; ` `} `   `// Utility function to create a new tree node ` `static` `Node newNode(``int` `data) ` `{ ` `    ``Node temp = ``new` `Node(); ` `    ``temp.data = data; ` `    ``temp.left = ``null``;` `    ``temp.right = ``null``; ` `    ``return` `temp; ` `} `   `// Driver program to test above functions ` `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``); `   `    ``int` `k = ``2``; ` `    ``int` `node = ``5``; `   `    ``// print kth ancestor of given node ` `    ``System.out.println(kthAncestor(root,``5``,k,node)); ` `}` `} `

## 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 newNode ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# function to generate array of ancestors ` `def` `generateArray(root, ancestors): `   `    ``# There will be no ancestor of root node ` `    ``ancestors[root.data] ``=` `-``1`   `    ``# level order traversal to ` `    ``# generate 1st ancestor ` `    ``q ``=` `[] ` `    ``q.append(root) `   `    ``while``(``len``(q)):` `        ``temp ``=` `q[``0``] ` `        ``q.pop(``0``) `   `        ``if` `(temp.left):` `            ``ancestors[temp.left.data] ``=` `temp.data ` `            ``q.append(temp.left) ` `    `  `        ``if` `(temp.right):` `            ``ancestors[temp.right.data] ``=` `temp.data ` `            ``q.append(temp.right) `   `# function to calculate Kth ancestor ` `def` `kthAncestor(root, n, k, node):` `    `  `    ``# create array to store 1st ancestors ` `    ``ancestors ``=` `[``0``] ``*` `(n ``+` `1``)`   `    ``# generate first ancestor array ` `    ``generateArray(root,ancestors) `   `    ``# variable to track record of number ` `    ``# of ancestors visited ` `    ``count ``=` `0`   `    ``while` `(node !``=` `-``1``) :` `        ``node ``=` `ancestors[node]` `        ``count ``+``=` `1` `        ``if``(count ``=``=` `k):` `            ``break` `            `  `    ``# print Kth ancestor ` `    ``return` `node` `                        `  `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``# Let us create binary tree shown ` `    ``# in above diagram ` `    ``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 ` `    ``print``(kthAncestor(root, ``5``, k, node))`   `# This code is contributed by ` `# SHUBHAMSINGH10`

## C#

 `/* C# program to calculate Kth ancestor of given node */` `using` `System;` `using` `System.Collections.Generic;`   `class` `GfG ` `{` `    `  `// A Binary Tree Node ` `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` `}`   `// function to generate array of ancestors ` `static` `void` `generateArray(Node root, ``int` `[]ancestors) ` `{ ` `    ``// There will be no ancestor of root node ` `    ``ancestors[root.data] = -1; `   `    ``// level order traversal to ` `    ``// generate 1st ancestor ` `    ``LinkedList q = ``new` `LinkedList (); ` `    ``q.AddLast(root); `   `    ``while``(q.Count != 0) ` `    ``{ ` `        ``Node temp = q.First.Value; ` `        ``q.RemoveFirst(); `   `        ``if` `(temp.left != ``null``) ` `        ``{ ` `            ``ancestors[temp.left.data] = temp.data; ` `            ``q.AddLast(temp.left); ` `        ``} `   `        ``if` `(temp.right != ``null``) ` `        ``{ ` `            ``ancestors[temp.right.data] = temp.data; ` `            ``q.AddLast(temp.right); ` `        ``} ` `    ``} ` `} `   `// function to calculate Kth ancestor ` `static` `int` `kthAncestor(Node root, ``int` `n, ``int` `k, ``int` `node) ` `{ ` `    ``// create array to store 1st ancestors ` `    ``int` `[]ancestors = ``new` `int``[n + 1]; `   `    ``// generate first ancestor array ` `    ``generateArray(root,ancestors); `   `    ``// variable to track record of number of ` `    ``// ancestors visited ` `    ``int` `count = 0; `   `    ``while` `(node != -1) ` `    ``{ ` `        ``node = ancestors[node]; ` `        ``count++; `   `        ``if``(count == k) ` `            ``break``; ` `    ``} `   `    ``// print Kth ancestor ` `    ``return` `node; ` `} `   `// Utility function to create a new tree node ` `static` `Node newNode(``int` `data) ` `{ ` `    ``Node temp = ``new` `Node(); ` `    ``temp.data = data; ` `    ``temp.left = ``null``;` `    ``temp.right = ``null``; ` `    ``return` `temp; ` `} `   `// Driver program to test above functions ` `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); `   `    ``int` `k = 2; ` `    ``int` `node = 5; `   `    ``// print kth ancestor of given node ` `    ``Console.WriteLine(kthAncestor(root,5,k,node)); ` `}` `} `   `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output

`1`

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 2: In this method first we will get an element whose ancestor has to be searched and from that node, we will decrement count one by one till we reach that ancestor node.
for example –

```consider the tree given below:-
(1)
/    \
(4)   (2)
/    \      \
(3)  (7)    (6)
\
(8)
Then check if k=0 if yes then return that element as an ancestor else climb a level up and reduce k (k = k-1).
Initially k = 2
First we search for 8 then,
at 8 => check if(k == 0) but k = 2 so k = k-1 => k = 2-1 = 1 and climb a level up i.e. at 7
at 7 => check if(k == 0) but k = 1 so k = k-1 => k = 1-1 = 0 and climb a level up i.e. at 4
at 4 => check if(k == 0) yes k = 0 return this node as ancestor.```

Implementation:

## C++14

 `// C++ program for finding ` `// kth ancestor of a particular node` `#include` `using` `namespace` `std;`   `// Structure for a node` `struct` `node{` `  ``int` `data;` `  ``struct` `node *left, *right;` `  ``node(``int` `x)` `  ``{` `    ``data = x;` `    ``left = right = NULL;` `  ``}` `};`   `// Program to find kth ancestor` `bool` `ancestor(``struct` `node* root, ``int` `item, ``int` `&k)` `{` `  ``if``(root == NULL)` `    ``return` `false``;` `  `  `  ``// Element whose ancestor is to be searched` `  ``if``(root->data == item)` `  ``{` `    ``//reduce count by 1` `    ``k = k-1;` `    ``return` `true``;` `  ``}` `  ``else` `  ``{` `    `  `    ``// Checking in left side` `    ``bool` `flag = ancestor(root->left,item,k);` `    ``if``(flag)` `    ``{` `      ``if``(k == 0)` `      ``{` `        `  `        ``// If count = 0 i.e. element is found` `        ``cout<<``"["``<data<<``"] "``;` `        ``return` `false``;` `      ``}` `      `  `      ``// if count !=0 i.e. this is not the ` `      ``// ancestor we are searching for ` `      ``// so decrement count` `      ``k = k-1;` `      ``return` `true``;` `    ``}`   `    ``// Similarly Checking in right side` `    ``bool` `flag2 = ancestor(root->right,item,k);` `    ``if``(flag2)` `    ``{` `      ``if``(k == 0)` `      ``{` `        ``cout<<``"["``<data<<``"] "``;` `        ``return` `false``;` `      ``}` `      ``k = k-1;` `      ``return` `true``;` `    ``}` `  ``}` `}`   `// Driver Code` `int` `main()` `{` `  ``struct` `node* root = ``new` `node(1);` `  ``root->left = ``new` `node(4);` `  ``root->left->right = ``new` `node(7);` `  ``root->left->left = ``new` `node(3);` `  ``root->left->right->left = ``new` `node(8);` `  ``root->right = ``new` `node(2);` `  ``root->right->right = ``new` `node(6);`   `  ``int` `item,k;` `  ``item = 3;` `  ``k = 1;` `  ``int` `loc = k;` `  ``bool` `flag =  ancestor(root,item,k);` `  ``if``(flag)` `       ``cout<<``"Ancestor doesn't exist\n"``;` `  ``else` `    ``cout<<``"is the "``<

## Java

 `// Java program for finding ` `// kth ancestor of a particular node` `import` `java.io.*;`   `class` `Node` `{` `    ``int` `data;` `    ``Node left, right;` `    `  `    ``Node(``int` `x)` `    ``{` `        ``this``.data = x;` `        ``this``.left = ``this``.right = ``null``;` `    ``}` `}`   `class` `GFG{` `    `  `static` `int` `k = ``1``;`   `static` `boolean` `ancestor(Node root, ``int` `item)` `{` `    ``if` `(root == ``null``)` `        ``return` `false``;` `    `  `    ``// Element whose ancestor is to be searched` `    ``if` `(root.data == item)` `    ``{` `        `  `        ``// Reduce count by 1` `        ``k = k-``1``;` `        ``return` `true``;` `    ``}` `    ``else` `    ``{` `    `  `        ``// Checking in left side` `        ``boolean` `flag = ancestor(root.left, item);` `        ``if` `(flag)` `        ``{` `            ``if` `(k == ``0``)` `            ``{` `            `  `                ``// If count = 0 i.e. element is found` `                ``System.out.print(``"["` `+ root.data + ``"] "``);` `                ``return` `false``;` `            ``}` `        `  `            ``// If count !=0 i.e. this is not the` `            ``// ancestor we are searching for` `            ``// so decrement count` `            ``k = k - ``1``;` `            ``return` `true``;` `        ``}` `    `  `        ``// Similarly Checking in right side` `        ``boolean` `flag2 = ancestor(root.right, item);` `        ``if` `(flag2)` `        ``{` `            ``if` `(k == ``0``)` `            ``{` `                ``System.out.print(``"["` `+ root.data + ``"] "``);` `                ``return` `false``;` `            ``}` `            ``k = k - ``1``;` `            ``return` `true``;` `        ``}` `    ``}` `    ``return` `false``;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``Node root = ``new` `Node(``1``);` `    ``root.left = ``new` `Node(``4``);` `    ``root.left.right = ``new` `Node(``7``);` `    ``root.left.left = ``new` `Node(``3``);` `    ``root.left.right.left = ``new` `Node(``8``);` `    ``root.right = ``new` `Node(``2``);` `    ``root.right.right = ``new` `Node(``6``);` `    `  `    ``int` `item = ``3``;` `    ``int` `loc = k;` `    ``boolean` `flag = ancestor(root, item);` `    `  `    ``if` `(flag)` `        ``System.out.println(``"Ancestor doesn't exist"``);` `    ``else` `        ``System.out.println(``"is the "` `+ (loc) +` `                           ``"th ancestor of ["` `+ ` `                           ``(item) + ``"]"``);` `}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Python3

 `# Python3 program for finding ` `# kth ancestor of a particular node` ` `  `# Structure for a node` `class` `node:` `    `  `    ``def` `__init__(``self``, data):` `        `  `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `        ``self``.data ``=` `data`   `# Program to find kth ancestor` `def` `ancestor(root, item):` `    `  `    ``global` `k` `    `  `    ``if` `(root ``=``=` `None``):` `        ``return` `False` `    `  `    ``# Element whose ancestor is` `    ``# to be searched` `    ``if` `(root.data ``=``=` `item):` `        `  `        ``# Reduce count by 1` `        ``k ``=` `k ``-` `1` `        ``return` `True` `  `  `    ``else``:` ` `  `        ``# Checking in left side` `        ``flag ``=` `ancestor(root.left, item);` `        `  `        ``if` `(flag):` `            ``if` `(k ``=``=` `0``):` `                `  `                ``# If count = 0 i.e. element is found` `                ``print``(``"["` `+` `str``(root.data) ``+` `"]"``, end ``=` `' '``)` `                ``return` `False` `        `  `            ``# If count !=0 i.e. this is not the ` `            ``# ancestor we are searching for ` `            ``# so decrement count` `            ``k ``=` `k ``-` `1` `            ``return` `True` `    `  `        ``# Similarly Checking in right side` `        ``flag2 ``=` `ancestor(root.right, item)` `        `  `        ``if` `(flag2):` `            ``if` `(k ``=``=` `0``):` `                ``print``(``"["` `+` `str``(root.data) ``+` `"]"``)` `                ``return` `False` `      `  `            ``k ``=` `k ``-` `1` `            ``return` `True`   `# Driver code` `if` `__name__``=``=``"__main__"``:` `    `  `    ``root ``=` `node(``1``)` `    ``root.left ``=` `node(``4``)` `    ``root.left.right ``=` `node(``7``)` `    ``root.left.left ``=` `node(``3``)` `    ``root.left.right.left ``=` `node(``8``)` `    ``root.right ``=` `node(``2``)` `    ``root.right.right ``=` `node(``6``)` `    `  `    ``item ``=` `3` `    ``k ``=` `1` `    ``loc ``=` `k` `    ``flag ``=` `ancestor(root, item)` `    `  `    ``if` `(flag):` `        ``print``(``"Ancestor doesn't exist"``)` `    ``else``:` `        ``print``(``"is the "` `+` `str``(loc) ``+` `              ``"th ancestor of ["` `+` `str``(item) ``+` `"]"``)` `     `  `# This code is contributed by rutvik_56`

## C#

 `// C# program for finding ` `// kth ancestor of a particular node` `using` `System;`     `// Structure for a node` `public` `class` `Node` `{` `    ``public` `int` `data;` `    ``public` `Node left, right;` `     `  `    ``public` `Node(``int` `x)` `    ``{` `        ``this``.data = x;` `        ``this``.left = ``this``.right = ``null``;` `    ``}` `}`   `class` `GFG{` `    `  `static` `int` `k = 1;`   `// Program to find kth ancestor` `static` `bool` `ancestor(Node root, ``int` `item)` `{` `    ``if` `(root == ``null``)` `        ``return` `false``;` `     `  `    ``// Element whose ancestor is ` `    ``// to be searched` `    ``if` `(root.data == item)` `    ``{` `        `  `        ``// Reduce count by 1` `        ``k = k - 1;` `        ``return` `true``;` `    ``}` `    ``else` `    ``{` `        `  `        ``// Checking in left side` `        ``bool` `flag = ancestor(root.left, item);` `        ``if` `(flag)` `        ``{` `            ``if` `(k == 0)` `            ``{` `                `  `                ``// If count = 0 i.e. element is found` `                ``Console.Write(``"["` `+ root.data + ``"] "``);` `                ``return` `false``;` `            ``}` `         `  `            ``// If count !=0 i.e. this is not the` `            ``// ancestor we are searching for` `            ``// so decrement count` `            ``k = k - 1;` `            ``return` `true``;` `        ``}` `     `  `        ``// Similarly Checking in right side` `        ``bool` `flag2 = ancestor(root.right, item);` `        ``if` `(flag2)` `        ``{` `            ``if` `(k == 0)` `            ``{` `                ``Console.Write(``"["` `+ root.data + ``"] "``);` `                ``return` `false``;` `            ``}` `            ``k = k - 1;` `            ``return` `true``;` `        ``}` `    ``}` `    ``return` `false``;` `}`   `// Driver code` `static` `public` `void` `Main()` `{` `    ``Node root = ``new` `Node(1);` `    ``root.left = ``new` `Node(4);` `    ``root.left.right = ``new` `Node(7);` `    ``root.left.left = ``new` `Node(3);` `    ``root.left.right.left = ``new` `Node(8);` `    ``root.right = ``new` `Node(2);` `    ``root.right.right = ``new` `Node(6);` `     `  `    ``int` `item = 3;` `    ``int` `loc = k;` `    ``bool` `flag = ancestor(root, item);` `     `  `    ``if` `(flag)` `        ``Console.WriteLine(``"Ancestor doesn't exist"``);` `    ``else` `        ``Console.WriteLine(``"is the "` `+ (loc) +` `                          ``"th ancestor of ["` `+` `                          ``(item) + ``"]"``);` `}` `}`   `// This code is contributed by patel2127`

## Javascript

 ``

Output

`[4] is the 1th ancestor of [3]`

Time Complexity: O(n)
Auxiliary Space: O(n)

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