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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.
#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.