103. Binary Tree Zigzag Level Order Traversal
Total Accepted: 64518 Total Submissions: 219636 Difficulty: Medium
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,null,null,15,7],
1 3 2 / 3 9 20 4 / 5 15 7
return its zigzag level order traversal as:
1 [ 2 [3], 3 [20,9], 4 [15,7] 5 ]
思路：
代码：
先层序遍历，然后将编号为奇数（从0开始）的vector中元素反转。
方法一：
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 queue<TreeNode*> q; 14 vector<vector<int>> res; 15 if(root==NULL){ 16 return res; 17 } 18 q.push(root); 19 q.push(NULL); 20 vector<int> temp; 21 bool even=true; 22 while(!q.empty()){ 23 TreeNode* cur=q.front(); 24 q.pop(); 25 if(cur){ 26 temp.push_back(cur->val); 27 if(cur->left) q.push(cur->left); 28 if(cur->right) q.push(cur->right); 29 } 30 else{ 31 even=!even; 32 res.push_back(temp); 33 if(q.size()>0){ 34 temp.resize(0); 35 q.push(NULL); 36 } 37 } 38 } 39 for(int i=0;i<res.size();i++){ 40 if(i%2){ 41 reverse(res[i].begin(),res[i].end()); 42 } 43 } 44 return res; 45 } 46 };
方法二：
1 class Solution { 2 public: 3 vector<vector<int>> zigzagLevelOrder(TreeNode* root) { 4 queue<TreeNode*> q; 5 vector<vector<int>> res; 6 if(root==NULL){ 7 return res; 8 } 9 q.push(root); 10 q.push(NULL); 11 vector<int> temp; 12 bool even=true; 13 while(!q.empty()){ 14 TreeNode* cur=q.front(); 15 q.pop(); 16 if(cur){ 17 if(even){ 18 temp.push_back(cur->val); 19 } 20 else{ 21 temp.insert(temp.begin(),cur->val); 22 } 23 if(cur->left) q.push(cur->left); 24 if(cur->right) q.push(cur->right); 25 } 26 else{ 27 even=!even; 28 res.push_back(temp); 29 if(q.size()>0){ 30 temp.resize(0); 31 q.push(NULL); 32 } 33 } 34 } 35 return res; 36 } 37 };