LeetCode in Net
103. Binary Tree Zigzag Level Order Traversal
Medium
Given the root of 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).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -100 <= Node.val <= 100
Solution
using System.Collections.Generic;
using LeetCodeNet.Com_github_leetcode;
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public IList<IList<int>> ZigzagLevelOrder(TreeNode root) {
var queue = new Queue<TreeNode>();
var results = new List<IList<int>>();
if (root == null) {
return results;
}
var level = new List<int>();
queue.Enqueue(root);
queue.Enqueue(null);
var d = false;
while (queue.Count > 0) {
var c = queue.Dequeue();
if (c == null) {
if (d) {
level.Reverse();
}
results.Add(level);
if (queue.Count == 0) {
break;
} else {
queue.Enqueue(null);
level = new List<int>();
d = !d;
}
} else {
level.Add((int)c.val);
if (c.left != null) {
queue.Enqueue(c.left);
}
if (c.right != null) {
queue.Enqueue(c.right);
}
}
}
return results;
}
}