LeetCode-in-Java
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
import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) {
return result;
}
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
q.add(null);
boolean zig = true;
LinkedList<Integer> level = new LinkedList<>();
while (!q.isEmpty()) {
TreeNode node = q.remove();
while (!q.isEmpty() && node != null) {
if (zig) {
level.add(node.val);
} else {
level.addFirst(node.val);
}
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node.right);
}
node = q.remove();
}
result.add(level);
zig = !zig;
level = new LinkedList<>();
if (!q.isEmpty()) {
q.add(null);
}
}
return result;
}
}