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Binary Tree Zigzag Level Order Traversal

Problem Description​

Given a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).

Examples​

Example 1:

LeetCode Problem - Binary Tree

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]

Constraints​

  • The number of nodes in the tree is in the range [0,2000][0, 2000].
  • −100≤Node.val≤100-100 \leq \text{Node.val} \leq 100

Solution for Binary Tree Problem​

Intuition And Approach​

To perform a zigzag level order traversal, we can use a breadth-first search (BFS) with a queue. We'll use a boolean flag to determine the direction of traversal at each level.

  1. Breadth-First Search (BFS): Traverse the tree level by level.
  2. Direction Toggle: Use a flag to toggle the direction of traversal for each level (left-to-right or right-to-left).

Code in Different Languages​

Written by @Vipullakum007
class Solution {
public int minDepth(TreeNode root) {
if (root == null) return 0;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int depth = 1;
while (!queue.isEmpty()) {
int levelSize = queue.size();
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
if (node.left == null && node.right == null) return depth;
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
depth++;
}
return depth;
}
}

Complexity Analysis​

  • Time Complexity: O(n)O(n) where n is the number of nodes in the binary tree.
  • Space Complexity: O(h)O(h) where h is the height of the binary tree.