LEETCODE: Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2025-12-02 15:32:05 发布

原创最新推荐文章于 2025-12-02 15:32:05 发布 · 282 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LEETCODE #算法 #二叉树 #遍历

LEETCODE 专栏收录该内容

153 篇文章

订阅专栏

本文介绍了一种算法，用于实现二叉树节点值的锯齿形层序遍历。即先从左到右遍历一层，接着从右到左遍历下一层，并交替进行。通过使用队列保存每一层的节点，再根据标志位判断是否需要反转当前层的节点值顺序。

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]
]

同上一篇，只是多做了一次reverse： http://blog.youkuaiyun.com/anyicheng2015/article/details/42264721

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int> > results;
        if(root == NULL) return results;
        queue<TreeNode *> onelevel;
        onelevel.push(root);
        bool shouldReverse = false;
        while(!onelevel.empty()) {
            // Get a level and get TreeNode of the next level.
            vector<int> valOfCurrentLevel;
            queue<TreeNode *> nextlevel;
            while(!onelevel.empty()) {
                TreeNode *front = onelevel.front();
                valOfCurrentLevel.push_back(front->val);
                if(front->left != NULL)
                    nextlevel.push(front->left);
                if(front->right != NULL)
                    nextlevel.push(front->right);
                onelevel.pop();
            }
            if(shouldReverse)
                reverse(valOfCurrentLevel.begin(), valOfCurrentLevel.end());
            results.push_back(valOfCurrentLevel);
            onelevel = nextlevel;
            shouldReverse = !shouldReverse;
        }
        return results;
    }
};