<LeetCode OJ> 103. Binary Tree Zigzag Level Order Traversal
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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.
分析:
这个问题和前面问题没有本质的区别吧?
<LeetCode OJ> 102. / 107. Binary Tree Level Order Traversal(I / II),
博客地址,http://blog.csdn.net/ebowtang/article/details/50719417
唯一的区别就是奇数层和偶数层的压入顺序相反而已,但是通过vector插入顺序的不同来达到目的似乎并没有达到题目的考察目的。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector< vector<int> > result; if(root==NULL) return result; queue<TreeNode*> que;//用来总是保存当层的节点 que.push(root); int level=1; //获取每一层的节点 while(!que.empty()) { int levelNum = que.size();//通过size来判断当层的结束 vector<int> curlevel; for(int i=0; i<levelNum; i++) { if(que.front()->left != NULL) //先获取该节点下一层的左右子,再获取该节点的元素,因为一旦压入必弹出,所以先处理左右子 que.push(que.front()->left); if(que.front()->right != NULL) que.push(que.front()->right); if(level&0x0001) //奇数层,或者if(level%2==1) curlevel.push_back(que.front()->val); else//偶数层 curlevel.insert(curlevel.begin(),que.front()->val); que.pop(); } level++; result.push_back(curlevel); } return result; }};
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原文地址:http://blog.csdn.net/ebowtang/article/details/50723062
原作者博客:http://blog.csdn.net/ebowtang
本博客LeetCode题解索引:http://blog.csdn.net/ebowtang/article/details/50668895
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