Reverse Level Order Traversal of a Binary Tree using Iteration
Visualization Player
Problem Statement
Examples
Solution
Algorithm Steps
Code
C
C++
Python
Java
JS
Go
Rust
Kotlin
Swift
TS
#include <stdio.h>
#include <stdlib.h>
typedef struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
} TreeNode;
TreeNode* createNode(int val) {
TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
node->val = val;
node->left = node->right = NULL;
return node;
}
int* reverseLevelOrder(TreeNode* root, int* returnSize) {
if (!root) {
*returnSize = 0;
return NULL;
}
TreeNode** queue = malloc(100 * sizeof(TreeNode*));
TreeNode** stack = malloc(100 * sizeof(TreeNode*));
int front = 0, rear = 0, top = -1;
queue[rear++] = root;
while (front < rear) {
TreeNode* node = queue[front++];
stack[++top] = node;
if (node->right) queue[rear++] = node->right;
if (node->left) queue[rear++] = node->left;
}
int* result = malloc(100 * sizeof(int));
int index = 0;
while (top >= 0) {
result[index++] = stack[top--]->val;
}
*returnSize = index;
free(queue);
free(stack);
return result;
}
int main() {
TreeNode* root = createNode(1);
root->left = createNode(2);
root->right = createNode(3);
root->left->left = createNode(4);
root->left->right = createNode(5);
int returnSize;
int* res = reverseLevelOrder(root, &returnSize);
printf("Reverse Level Order Traversal: ");
for (int i = 0; i < returnSize; i++) {
printf("%d ", res[i]);
}
printf("\n");
free(res);
return 0;
}
Comments
Loading comments...