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Copy path103.二叉树的锯齿形层次遍历.java
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103.二叉树的锯齿形层次遍历.java
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/*
* @lc app=leetcode.cn id=103 lang=java
*
* [103] 二叉树的锯齿形层次遍历
*
* https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/description/
*
* algorithms
* Medium (50.70%)
* Likes: 90
* Dislikes: 0
* Total Accepted: 17.1K
* Total Submissions: 33.2K
* Testcase Example: '[3,9,20,null,null,15,7]'
*
* 给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
*
* 例如:
* 给定二叉树 [3,9,20,null,null,15,7],
*
* 3
* / \
* 9 20
* / \
* 15 7
*
*
* 返回锯齿形层次遍历如下:
*
* [
* [3],
* [20,9],
* [15,7]
* ]
*
*
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) {
return result;
}
zigzagLevelOrder(result, root);
return result;
}
private void zigzagLevelOrder(List<List<Integer>> result, TreeNode root) {
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int level = 1;
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> row = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
assert node != null;
row.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
if (level % 2 == 0) {
Collections.reverse(row);
}
result.add(row);
level++;
}
}
}