[LeetCode]Binary Tree Zigzag Level Order Traversal-优快云博客

本文介绍了一种解决二叉树节点值的锯齿形层序遍历问题的方法，即从左到右再从右到左交替进行的层序遍历。通过使用BFS（宽度优先搜索）策略实现这一过程，并详细展示了如何根据当前层级来决定是否需要反转该层节点值的顺序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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]
]

标准BFS，层序遍历。

/**
 * Definition for a binary tree node.
 * 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>> res;
        vector<int> traversal;
        int level = 0;
        queue<TreeNode*> que;
        if(root==NULL)  return res;
        que.push(root);
        while(!que.empty()){
            int len = que.size();
            traversal.clear();
            for(int i=0; i<len; ++i){//BFS搜索每次把一整层的内容出队列
                TreeNode* FNode = que.front();
                traversal.push_back(FNode->val);
                if(FNode->left) que.push(FNode->left);
                if(FNode->right) que.push(FNode->right);
                que.pop();
            }
            if(level%2==0){
                res.push_back(traversal);
            }
            if(level%2==1){
                res.push_back(vector<int>(traversal.rbegin(),traversal.rend()));//翻转vector数组
            }
            level++;
        }
        return res;
    }
};