# Kth node in Diagonal Traversal of Binary Tree in C++

In this tutorial, we are going to write a program that finds the k-th node in the diagonal traversal of a binary tree.

Let's see the steps to solve the problem.

• Initialise the binary tree with some sample data.
• Initialise the number k.
• Traverse the binary tree diagonally using the data structure queue.
• Decrement the value of k on each node.
• Return the node when k becomes 0.
• Return -1 if there is no such node.

## Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node *left, *right;
};
Node* getNewNode(int data) {
Node* node = (Node*)malloc(sizeof(Node));
node->data = data;
node->left = node->right = NULL;
return node;
}
int findDiagonalKthElement(Node* root, int k) {
if (root == NULL || k == 0) {
return -1;
}
int result = -1;
queue<Node*> q;
q.push(root);
q.push(NULL);
while (!q.empty()) {
Node* temp = q.front();
q.pop();
if (temp == NULL) {
if (q.empty()) {
if (k == 0) {
return result;
}else {
break;
}
}
q.push(NULL);
}else {
while (temp) {
if (k == 0) {
return result;
}
k--;
result = temp->data;
if (temp->left) {
q.push(temp->left);
}
temp = temp->right;
}
}
}
return -1;
}
int main() {
Node* root = getNewNode(10);
root->left = getNewNode(5);
root->right = getNewNode(56);
root->left->left = getNewNode(3);
root->left->right = getNewNode(22);
root->right->right = getNewNode(34);
root->right->right->left = getNewNode(45);
root->left->right->left = getNewNode(67);
root->left->right->right = getNewNode(100);
int k = 9;
cout << findDiagonalKthElement(root, k) << endl;
return 0;
}

## Output

If you run the above code, then you will get the following result.

67

## Conclusion

If you have any queries in the tutorial, mention them in the comment section.

Updated on: 09-Apr-2021

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