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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},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
1 class Solution { 2 public: 3 vector<vector<int>> zigzagLevelOrder(TreeNode* root) { 4 vector<vector<int> >res; 5 if(root==NULL) return res; 6 7 vector<int> level; 8 queue<TreeNode* > q; 9 q.push(root); 10 11 int cnt=1; 12 int curcnt=0; 13 14 bool order=true; 15 16 while(!q.empty()) 17 { 18 TreeNode *node=q.front(); 19 q.pop(); 20 level.push_back(node->val); 21 if(node->left) 22 { 23 q.push(node->left); 24 curcnt++; 25 } 26 if(node->right) 27 { 28 q.push(node->right); 29 curcnt++; 30 } 31 32 cnt--; 33 if(cnt==0) 34 { 35 if(order) 36 { 37 res.push_back(level); 38 }else{ 39 reverse(level.begin(),level.end()); 40 res.push_back(level); 41 } 42 43 order=!order; 44 level.clear(); 45 cnt=curcnt; 46 curcnt=0; 47 } 48 } 49 return res; 50 } 51 };
【leetcode】Binary Tree Zigzag Level Order Traversal
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原文地址：http://www.cnblogs.com/jawiezhu/p/4557465.html
