LeetCode Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2025-08-11 15:06:51 发布

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最新推荐文章于 2025-08-11 15:06:51 发布

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分类专栏： LeetCode刷题文章标签： leetcode 二叉树遍历

本文链接：https://blog.youkuaiyun.com/sun_wangdong/article/details/50195673

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本文介绍了一种特殊的二叉树遍历方法——之字形遍历，并提供了一个具体的实现示例。该方法首先从根节点开始，随后每一层的节点值按从左到右或从右到左的顺序收集，交替进行。

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题目：

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

For example:
Given binary tree {3,9,20,#,#,15,7},

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

题意：

就是给定一棵二叉树，然后按照之字形来输出，其实就是层次遍历，只不过遇到偶数层，就换一下顺序即可，换汤不换药的题，和我之前的那两篇博客的方法基本一样。

public class Solution 
{
    public List<List<Integer>> zigzagLevelOrder(TreeNode root)
	{
		List<List<Integer>> list = new ArrayList<List<Integer>>();
		
		LinkedList<TreeNode> node = new LinkedList<TreeNode>();
		if(root == null)
			return list;
		else
		{
		    node.add(root);
		    int length = node.size();
		    int i = 1;
		    while(!node.isEmpty())
		    {
		    	List<Integer> nodes = new ArrayList<Integer>();
		    	while(length-- > 0)
		    	{
		    		int value = node.peek().val;
		    		TreeNode n = node.peek();
		    		node.poll();
		    		nodes.add(value);
		    		if(n.left != null)
		    			node.add(n.left);
		    		if(n.right != null)
		    			node.add(n.right);
		    	}
		    	length = node.size();
		    	if(i % 2 == 0)
		    		Collections.reverse(nodes);
		    	list.add(nodes);
		    	i++;
		    }
		    return list;
		}
	}
}