文章目录
一、题目
Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1:
Input: root = 3,9,20,null,null,15,7
Output: \[3,20,9,15,7]
Example 2:
Input: root = 1
Output: \[1]
Example 3:
Input: root = \[\]
Output: \[\]
Constraints:
The number of nodes in the tree is in the range 0, 2000.
-100 <= Node.val <= 100
二、题解
cpp
/**
* 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>> 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;
}
};