binary-tree-zigzag-level-order-traversal——二叉树分层输出
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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.
OJ‘s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1 / 2 3 / 4 5The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
PS:二叉树分层输出,BFS并且记录翻转情况
1 /** 2 * Definition for binary tree 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==NULL) return res; 15 queue<TreeNode*> q; 16 q.push(root); 17 bool reverse=false; 18 while(!q.empty()){ 19 vector<int> v; 20 int size=q.size(); 21 for(int i=0;i<size;++i){ 22 TreeNode *cur=q.front(); 23 q.pop(); 24 v.push_back(cur->val); 25 if(cur->left!=NULL) q.push(cur->left); 26 if(cur->right!=NULL) q.push(cur->right); 27 } 28 if(reverse){ 29 30 vector<int> tmp; 31 for(int i=v.size()-1;i>=0;--i){ 32 tmp.push_back(v[i]); 33 } 34 res.push_back(tmp); 35 }else{ 36 res.push_back(v); 37 } 38 reverse=!reverse; 39 } 40 return res; 41 } 42 };
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