K-th ancestor of a node in Binary Tree

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. 

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 <iostream>
#include <queue>
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<Node*> 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<<kthAncestor(root,5,k,node);
    return 0;
}

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<Node> q = new LinkedList<Node> ();
    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<Node> q = new LinkedList<Node> ();
    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

<script>

/* JavaScript program to calculate
Kth ancestor of given node */

// A Binary Tree Node
class Node
{
  constructor()
  {
    this.data = 0;
    this.left = null;
    this.right = null;
  }
}

// function to generate array of ancestors
function generateArray(root, ancestors)
{
    // There will be no ancestor of root node
    ancestors[root.data] = -1;

    // level order traversal to
    // generate 1st ancestor
    var q = [];
    q.push(root);

    while(q.length != 0)
    {
        var temp = q[0];
        q.shift();

        if (temp.left != null)
        {
            ancestors[temp.left.data] = temp.data;
            q.push(temp.left);
        }

        if (temp.right != null)
        {
            ancestors[temp.right.data] = temp.data;
            q.push(temp.right);
        }
    }
}

// function to calculate Kth ancestor
function kthAncestor(root, n, k, node)
{
    // create array to store 1st ancestors
    var ancestors = Array(n+1).fill(0);

    // generate first ancestor array
    generateArray(root,ancestors);

    // variable to track record of number of
    // ancestors visited
    var 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
function newNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Driver program to test above functions
// Let us create binary tree shown in above diagram
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
var k = 2;
var node = 5;
// print kth ancestor of given node
document.write(kthAncestor(root,5,k,node));


</script>

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<bits/stdc++.h>
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<<"["<<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,k);
    if(flag2)
    {
      if(k == 0)
      {
        cout<<"["<<root->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 "<<loc<<"th ancestor of ["<<
                                   item<<"]"<<endl;
  return 0;
}

// This code is contributed by Sanjeev Yadav.

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

<script>

class Node
{
    constructor(x)
    {
        this.data=x;
        this.left = this.right = null;
    }
}

function ancestor(root, item)
{
    if (root == null)
    {
        return false;
    }

    if (root.data == item)
    {
        k = k - 1;
        return true;
    }

    else
    {
        let flag = ancestor(root.left, item);
        if (flag)
        {
            if (k == 0)
            {
                document.write("[" + (root.data) + "] ");
                return false;
            }
            k = k - 1;
            return true;
        }
        let flag2 = ancestor(root.right, item);
        if (flag2)
        {
            if (k == 0)
            {
                document.write("[" + (root.data) + "] ");
                return false;
            }
            k = k - 1;
            return true;
        }

    }
}

let 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)
let item = 3
    let k = 1
   let loc = k
   let flag = ancestor(root, item)

    if (flag)
        document.write("Ancestor doesn't exist")
    else
        document.write("is the " + (loc) +
              "th ancestor of [" + (item) + "]")

// This code is contributed by rag2127

</script>

Output

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

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

Method 3: Iterative Approach

The basic idea behind the iterative approach is to traverse the binary tree from the root node and keep track of the path from the root to the target node using a stack. Once we find the target node, we pop elements from the stack and add their values to a vector until we reach the kth ancestor or the stack becomes empty

Follow the Steps below to implement the above idea:

1. Initialize a stack to keep track of the path from the root to the target node, and a vector to store the ancestors.
2. Traverse the binary tree from the root node using a while loop.
3. If the current node is not NULL, push it onto the stack and move to its left child.
4. If the current node is NULL, pop the top element from the stack. If the top element is the target node, break out of the loop. Otherwise, move to its right child.
5. If the target node is not found, return -1.
6. Pop elements from the stack and add their values to the ancestors vector until we reach the kth ancestor or the stack becomes empty.
7. If the stack becomes empty before we reach the kth ancestor, return -1.
8. Return the value of the kth ancestor

Below is the implementation of the above approach:

C++

// C++ code to implement the iterative approach
#include <iostream>
#include <stack>
#include <vector>
using namespace std;

// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

// Function to find the kth ancestor of the given node using iterative approach
int kthAncestor(TreeNode* root, int node, int k) {
    // Initialize a stack to keep track of the path from the root to the target node
    stack<TreeNode*> s;
    vector<int> ancestors;
    bool found = false;

    // Traverse the binary tree from the root node
    while (root != NULL || !s.empty()) {
        // If the current node is not NULL, push it onto the stack and move to its left child
        if (root != NULL) {
            s.push(root);
            root = root->left;
        }
        // If the current node is NULL, pop the top element from the stack
        // If the top element is the target node, break out of the loop
        // Otherwise, move to its right child
        else {
            TreeNode* temp = s.top();
            s.pop();
            if (temp->val == node) {
                found = true;
                break;
            }
            if (temp->right != NULL) {
                root = temp->right;
            }
        }
    }

    // If the target node is not found, return -1
    if (!found) {
        return -1;
    }

    // Pop elements from the stack and add their values to the ancestors vector
    // until we reach the kth ancestor or the stack becomes empty
    while (!s.empty() && k > 0) {
        TreeNode* temp = s.top();
        s.pop();
        ancestors.push_back(temp->val);
        k--;
    }

    // If the stack becomes empty before we reach the kth ancestor, return -1
    if (k > 0) {
        return -1;
    }

    // Return the value of the kth ancestor
    return ancestors.back();
}

// Driver code
int main() {
    /*
        Example tree:
               1
             /   \
            2     3
           / \
          4   5
    */
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);

    int node = 4;
    int k = 2;

    int kthAncestorVal = kthAncestor(root, node, k);

    cout << "The " << k << "th ancestor of node " << node << " is " << kthAncestorVal << endl;

    return 0;
}
// This code is contributed by Veerendra_Singh_Rajpoot

Python3

# Python code to implement the iterative approach

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

# Function to find the kth ancestor of
# the given node using iterative approach
def kthAncestor(root, node, k):

    # Initialize a stack to keep track of
    # the path from the root to the target node
    s = []
    ancestors = []
    found = False

    # Traverse the binary tree from the root node
    while root or s:
        # If the current node is not None, push it onto the stack and move to its left child
        if root:
            s.append(root)
            root = root.left
        # If the current node is None, pop the top element from the stack
        # If the top element is the target node, break out of the loop
        # Otherwise, move to its right child
        else:
            temp = s.pop()
            if temp.val == node:
                found = True
                break
            if temp.right:
                root = temp.right

    # If the target node is not found, return -1
    if not found:
        return -1

    # Pop elements from the stack and add their values to the ancestors vector
    # until we reach the kth ancestor or the stack becomes empty
    while s and k > 0:
        temp = s.pop()
        ancestors.append(temp.val)
        k -= 1

    # If the stack becomes empty before we reach the kth ancestor, return -1
    if k > 0:
        return -1

    # Return the value of the kth ancestor
    return ancestors[-1]

# Driver code
if __name__ == '__main__':
    '''
        Example tree:
               1
             /   \
            2     3
           / \
          4   5
    '''
    root = TreeNode(1)
    root.left = TreeNode(2)
    root.right = TreeNode(3)
    root.left.left = TreeNode(4)
    root.left.right = TreeNode(5)

    node = 4
    k = 2

    kthAncestorVal = kthAncestor(root, node, k)

    print("The", k, "th ancestor of node", node, "is", kthAncestorVal)

# This code is contributed by rishabmalhdijo

Output

The 2th ancestor of node 4 is 1

Time Complexity: O(N) , where N is the number of nodes in the binary tree. This is because we need to traverse the entire tree in the worst case to find the target node and the ancestors.

Space Complexity: O(N) , where N is the number of nodes in the binary tree. This is because we are using a stack to keep track of the path from the root to the target node, and in the worst case, the entire path could be stored in the stack

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.