LeetCode103. Binary Tree Zigzag Level Order Traversal

本文介绍: 【代码】Lee tCo de103. Bin a ry Tre e Zi gz a g Le v el Or de r Tr aver sal。

文章 目录

- 一、题目
- 二、题解

一、题目

Gi v e n th e root o f a binary tree, return th e zi gza g level order tr aver sal o f i ts node s’ values. (i.e., f ro m left t o right, then right t o left f o r the next lev el and a lte r nate between).

Ex ample 1:

Inpu t: root = [3,9,20,null,null,15,7]
Out put: [[3],[20,9],[15,7]]
Example 2:

Inpu t: root = [1]
Output: [[1]]
Example 3:

Input: root = []
Output: []

Con s train ts:

The number o f no des in the tree i s in the range [0, 2000].
-100 <= Node.val <= 100

二、题解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int&gt;&gt; zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        queue<TreeNode*> q;
        if(!root) return res;
        q.push(root);
        int level = 0;
        while(!q.empty()){
            int size = q.size();
            vector<int> tmp;
            while(size--){
                TreeNode* t = q.front();
                q.pop();
                tmp.push_back(t->val);
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            if(level % 2 == 1){
                reverse(tmp.begin(),tmp.end());
                res.push_back(tmp);
            }
            else res.push_back(tmp);
            level++;
        }
        return res;
    }
};