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In this problem, we are given a binary tree and a key **k**. We need to find out all the ancestors of **k**.

For example, if k = 7 and the given tree is -

Then, the ancestors of 7 will be 4,2 and 1.

You can simply solve this problem using standard traversal techniques like:

- Breadth First Search
- Depth First Search
- Level order traversal

The idea is to do that traversal recursively and return 1 when you reach the given node K and return 0 for all other cases. What this does is that: All nodes coming before the given node K get the return value of 1 and if we get 1, it means the current node is an ancestor.

This is a simple approach which you can take to use existing knowledge of BFS or DFS.

We will easily solve this problem using recusion:

- Let the recursive function be
**printAncestor(node, k)**, which will return a boolean (true/false). - We will start from the root node with k=7. The function
**printAncestor(root, k)**will be executed and it will recursively call its left and right nodes until**k**i.e 7 is found in that node, or till node = NULL. - We will return
**true**if**k**will be found in the node. - We will print a node whenever
**true**is returned from its left or right nodes. (That means the current node is the key's ancestor).

## Code -

The following is the code to the above problem in C++

```
#include<bits/stdc++.h>
using namespace std;
/* A binary tree node having data,
pointer to left child and
a pointer to right child */
class node
{
public:
int data;
node* left;
node* right;
/* Constructor that allocates a new node with the
given data and NULL left and right pointers. */
node(int data)
{
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
/* If target is present in any node, print its ancestor and return true. Else, return false*/
bool printAncestors(struct node *root, int target)
{
/* base cases */
if (root == NULL)
return false;
if (root->data == target)
return true;
/* If target is present in either left or right subtree of this node, then print this node */
if ( printAncestors(root->left, target) ||
printAncestors(root->right, target) )
{
cout << root->data << " ";
return true;
}
/* Else return false */
return false;
}
/* Driver code*/
int main()
{
/* Construct the following binary tree
1
/ \
2 3
/ / \
4 5 6
/
7
*/
node *root = new node(1);
root->left = new node(2);
root->right = new node(3);
root->left->left = new node(4);
root->right->left = new node(5);
root->right->right = new node(6);
root->left->left->left = new node(7);
printAncestors(root, 7);
return 0;
}
```

## Workflow

Let us walkthrough the problem with our given example, where k = 7 and the following binary tree is given -

- The function printAncestor(root (1), 7)
*(the number in the bracket denotes the data stored at that particular node)*will be called initially, which will recusrsively call its left and right subtrees until we get the function printAncestor(node (7), 7). - printAncestor(node (7), 7) will return
**true**in its parent node (4), hence the value 4 will be printed, and again we will return**true**. - node (2) will receive
**true**from its child, so it will print 2 and return**true**. This process is continued till the root node. Therefore, 4 2 1 is printed as the final answer.

The following graph depicts the recursion route for this problem -

## Complexity

**Time complexity: O(n)
Space complexity: O(n)**, where n is the number of nodes.

Hence, we have figured out how to find the ancestors of any given node.