# Print ancestors of a given binary tree node without recursion

Given a Binary Tree and a key, write a function that prints all the ancestors of the key in the given binary tree.

For example, consider the following Binary Tree

```            1
/       \
2         3
/   \     /   \
4     5    6    7
/       \       /
8         9     10  ```

Following are different input keys and their ancestors in the above tree

```Input Key    List of Ancestors
-------------------------
1
2            1
3            1
4            2 1
5            2 1
6            3 1
7            3 1
8            4 2 1
9            5 2 1
10            7 3 1
```

Recursive solution for this problem is discussed here.
It is clear that we need to use a stack based iterative traversal of the Binary Tree. The idea is to have all ancestors in stack when we reach the node with given key. Once we reach the key, all we have to do is, print contents of stack.
How to get all ancestors in stack when we reach the given node? We can traverse all nodes in Postorder way. If we take a closer look at the recursive postorder traversal, we can easily observe that, when recursive function is called for a node, the recursion call stack contains ancestors of the node. So idea is do iterative Postorder traversal and stop the traversal when we reach the desired node.

Following is implementation of the above approach.

## C

```// C program to print all ancestors of a given key
#include <stdio.h>
#include <stdlib.h>

// Maximum stack size
#define MAX_SIZE 100

// Structure for a tree node
struct Node
{
int data;
struct Node *left, *right;
};

// Structure for Stack
struct Stack
{
int size;
int top;
struct Node* *array;
};

// A utility function to create a new tree node
struct Node* newNode(int data)
{
struct Node* node = (struct Node*) malloc(sizeof(struct Node));
node->data = data;
node->left = node->right = NULL;
return node;
}

// A utility function to create a stack of given size
struct Stack* createStack(int size)
{
struct Stack* stack = (struct Stack*) malloc(sizeof(struct Stack));
stack->size = size;
stack->top = -1;
stack->array = (struct Node**) malloc(stack->size * sizeof(struct Node*));
return stack;
}

// BASIC OPERATIONS OF STACK
int isFull(struct Stack* stack)
{
return ((stack->top + 1) == stack->size);
}
int isEmpty(struct Stack* stack)
{
return stack->top == -1;
}
void push(struct Stack* stack, struct Node* node)
{
if (isFull(stack))
return;
stack->array[++stack->top] = node;
}
struct Node* pop(struct Stack* stack)
{
if (isEmpty(stack))
return NULL;
return stack->array[stack->top--];
}
struct Node* peek(struct Stack* stack)
{
if (isEmpty(stack))
return NULL;
return stack->array[stack->top];
}

// Iterative Function to print all ancestors of a given key
void printAncestors(struct Node *root, int key)
{
if (root == NULL) return;

// Create a stack to hold ancestors
struct Stack* stack = createStack(MAX_SIZE);

// Traverse the complete tree in postorder way till we find the key
while (1)
{
// Traverse the left side. While traversing, push the nodes into
// the stack so that their right subtrees can be traversed later
while (root && root->data != key)
{
push(stack, root);   // push current node
root = root->left;  // move to next node
}

// If the node whose ancestors are to be printed is found,
// then break the while loop.
if (root && root->data == key)
break;

// Check if right sub-tree exists for the node at top
// If not then pop that node because we don't need this
// node any more.
if (peek(stack)->right == NULL)
{
root = pop(stack);

// If the popped node is right child of top, then remove the top
// as well. Left child of the top must have processed before.
// Consider the following tree for example and key = 3.  If we
// remove the following loop, the program will go in an
// infinite loop after reaching 5.
//          1
//        /   \
//       2     3
//         \
//           4
//             \
//              5
while (!isEmpty(stack) && peek(stack)->right == root)
root = pop(stack);
}

// if stack is not empty then simply set the root as right child
// of top and start traversing right sub-tree.
root = isEmpty(stack)? NULL: peek(stack)->right;
}

// If stack is not empty, print contents of stack
// Here assumption is that the key is there in tree
while (!isEmpty(stack))
printf("%d ", pop(stack)->data);
}

// Driver program to test above functions
int main()
{
// Let us construct a binary tree
struct Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->left->left->left = newNode(8);
root->left->right->right = newNode(9);
root->right->right->left = newNode(10);

printf("Following are all keys and their ancestors\n");
for (int key = 1; key <= 10; key++)
{
printf("%d: ", key);
printAncestors(root, key);
printf("\n");
}

getchar();
return 0;
}
```

## C++

```// C++ program to print all ancestors of a given key
#include <bits/stdc++.h>
using namespace std;

// Structure for a tree node
struct Node
{
int data;
struct Node *left, *right;
};

// A utility function to create a new tree node
struct Node* newNode(int data)
{
struct Node* node = (struct Node*) malloc(sizeof(struct Node));
node->data = data;
node->left = node->right = NULL;
return node;
}

// Iterative Function to print all ancestors of a given key
void printAncestors(struct Node *root, int key)
{
if (root == NULL) return;

// Create a stack to hold ancestors
stack<struct Node* > st;

// Traverse the complete tree in postorder way till we find the key
while (1)
{
// Traverse the left side. While traversing, push the nodes into
// the stack so that their right subtrees can be traversed later
while (root && root->data != key)
{
st.push(root);   // push current node
root = root->left;  // move to next node
}

// If the node whose ancestors are to be printed is found,
// then break the while loop.
if (root && root->data == key)
break;

// Check if right sub-tree exists for the node at top
// If not then pop that node because we don't need this
// node any more.
if (st.top()->right == NULL)
{
root = st.top();
st.pop();

// If the popped node is right child of top, then remove the top
// as well. Left child of the top must have processed before.

while (!st.empty() && st.top()->right == root)
{root = st.top();
st.pop();
}
}

// if stack is not empty then simply set the root as right child
// of top and start traversing right sub-tree.
root = st.empty()? NULL: st.top()->right;
}

// If stack is not empty, print contents of stack
// Here assumption is that the key is there in tree
while (!st.empty())
{
cout<<st.top()->data<<" ";
st.pop();
}
}

// Driver program to test above functions
int main()
{
// Let us construct a binary tree
struct Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->left->left->left = newNode(8);
root->left->right->right = newNode(9);
root->right->right->left = newNode(10);

cout<<"Following are all keys and their ancestors"<<endl;
for (int key = 1; key <= 10; key++)
{
cout<<key<<":"<<" ";
printAncestors(root, key);
cout<<endl;
}

return 0;
}
// This code is contributed by Gautam Singh
```

## Java

```// Java program to print all ancestors of a given key
import java.util.Stack;

public class GFG
{
// Class for a tree node
static class Node
{
int data;
Node left,right;

// constructor to create Node
// left and right are by default null
Node(int data)
{
this.data = data;
}
}

// Iterative Function to print all ancestors of a given key
static void printAncestors(Node root,int key)
{
if(root == null)
return;

// Create a stack to hold ancestors
Stack<Node> st = new Stack<>();

// Traverse the complete tree in postorder way till we find the key
while(true)
{

// Traverse the left side. While traversing, push the nodes into
// the stack so that their right subtrees can be traversed later
while(root != null && root.data != key)
{
st.push(root);   // push current node
root = root.left;   // move to next node
}

// If the node whose ancestors are to be printed is found,
// then break the while loop.
if(root != null && root.data == key)
break;

// Check if right sub-tree exists for the node at top
// If not then pop that node because we don't need this
// node any more.
if(st.peek().right == null)
{
root =st.peek();
st.pop();

// If the popped node is right child of top, then remove the top
// as well. Left child of the top must have processed before.
while( st.empty() == false && st.peek().right == root)
{
root = st.peek();
st.pop();
}
}

// if stack is not empty then simply set the root as right child
// of top and start traversing right sub-tree.
root = st.empty() ? null : st.peek().right;
}

// If stack is not empty, print contents of stack
// Here assumption is that the key is there in tree
while( !st.empty() )
{
System.out.print(st.peek().data+" ");
st.pop();
}
}

// Driver program to test above functions
public static void main(String[] args)
{
// Let us construct a binary tree
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.right.right = new Node(9);
root.right.right.left = new Node(10);

System.out.println("Following are all keys and their ancestors");
for(int key = 1;key <= 10;key++)
{
System.out.print(key+": ");
printAncestors(root, key);
System.out.println();
}
}

}
//This code is Contributed by Sumit Ghosh
```

Output:
```Following are all keys and their ancestors
1:
2: 1
3: 1
4: 2 1
5: 2 1
6: 3 1
7: 3 1
8: 4 2 1
9: 5 2 1
10: 7 3 1```

Exercise
Note that the above solution assumes that the given key is present in the given Binary Tree. It may go in infinite loop if key is not present. Extend the above solution to work even when the key is not present in tree.

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