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).
Solution
由于是 /(zigzag/) 的形式，刚开始的思路是用 /(stack/) 来维护反序，但是不太自然；不如直接用双端队列来模拟(因为两头都可以 /(pop,push/))
点击查看代码
/**
* 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 {
private:
vector<vector<int>> ans;
deque<TreeNode*> dq;
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
if(!root) return ans;
dq.push_back(root);
int fg = false;
// 1: right -> left;
// 0: left -> right;
while(!dq.empty()){
vector<int> tmp;
int sz = dq.size();
while(sz--){
if(!fg){
// left -> right
auto f = dq.front(); dq.pop_front();
if(f->left)dq.push_back(f->left);
if(f->right)dq.push_back(f->right);
tmp.push_back(f->val);
}
else{
// right -> left
auto f = dq.back(); dq.pop_back();
if(f->right)dq.push_front(f->right);
if(f->left)dq.push_front(f->left);
tmp.push_back(f->val);
}
}
ans.push_back(tmp);
fg = !fg;
}
return ans;
}
};
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