LeetCode 103. Binary Tree Zigzag Level Order Traversal

Posted 2020-06-17

思路：维护两个stack<TreeNode*> stk1, stk2，stk1存放当前层的节点，stk2存放下一层的节点。

　　　bool right表示下一层的顺序，如果为True则下一层为从右到左，因为stk2是stack，所以stk2先存cur->left，然后存cur->right。

　　   当stk1为空，即当前层的节点遍历完时，将stk1、stk2之间swap，即永远对stk1进行处理，并把当前层生成的vector level压入最终的vector res中，注意level要clear且　            right要取反

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
13         vector<vector<int>> res;
14         if(!root) return res;
15         
16         vector<int> level;
17         stack<TreeNode* > stk1, stk2;
18         
19         stk1.push(root);
20         bool right = true;
21         
22         while( !stk1.empty() ){
23             TreeNode* cur = stk1.top();
24             stk1.pop();
25             level.push_back(cur->val);
26             
27             if(right){
28                 if(cur->left) stk2.push(cur->left);
29                 if(cur->right) stk2.push(cur->right);
30             }
31             else{
32                 if(cur->right) stk2.push(cur->right);
33                 if(cur->left) stk2.push(cur->left);
34             }
35             
36             if(stk1.empty()){
37                 res.push_back(level);
38                 level.clear();
39                 stk1.swap(stk2);
40                 right = !right;
41             }
42         }
43         
44         return res;
45         
46     }
47 };