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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] ]
简单的bfs而已，不过在bfs的过程中应该注意将相应的数组reverse一下，其实都到最终结果之后在隔行reverse也是可以的，下面给出非递归的版本，用递归同样也很好实现：
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 struct Node 12 { 13 TreeNode * node; 14 int level; 15 Node(){} 16 Node(TreeNode * n, int lv) 17 : node(n), level(lv){} 18 }; 19 public: 20 vector<vector<int>> zigzagLevelOrder(TreeNode* root) { 21 vector<vector<int>> ret; 22 if(!root) return ret; 23 vector<int> tmp; 24 int dep = 0; 25 queue<Node>q; 26 q.push(Node(root, 0)); 27 while(!q.empty()){ 28 Node tmpNode = q.front(); //非递归的使用bfs，借助队列特性 29 if(tmpNode.node->left) 30 q.push(Node(tmpNode.node->left, tmpNode.level + 1)); 31 if(tmpNode.node->right) 32 q.push(Node(tmpNode.node->right, tmpNode.level + 1)); 33 if(dep < tmpNode.level){ 34 if(dep % 2){ 35 reverse(tmp.begin(), tmp.end()); 36 } 37 ret.push_back(tmp); 38 tmp.clear(); 39 dep = tmpNode.level; 40 } 41 tmp.push_back(tmpNode.node->val); 42 q.pop(); 43 } 44 if(dep % 2){ 45 reverse(tmp.begin(), tmp.end()); 46 } 47 ret.push_back(tmp); 48 return ret; 49 } 50 };
LeetCode OJ：Binary Tree Zigzag Level Order Traversal（折叠二叉树遍历）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4910514.html
