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Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree
Given binary tree
{3,9,20,#,#,15,7}, 3
/ \
9 20
/ \
15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
confused what
» Solve this problem"{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.这题就是BFS,不论是图还是树的一种基本遍历的方法，应该掌握，比较简单，我用了2个queue来存，其实不需要，用一个count来计数每一层的node的个数，我使用了一个queue在level order traversal ii里面。
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| /** | |
| * Definition for binary tree | |
| * struct TreeNode { | |
| * int val; | |
| * TreeNode *left; | |
| * TreeNode *right; | |
| * TreeNode(int x) : val(x), left(NULL), right(NULL) {} | |
| * }; | |
| */ | |
| class Solution { | |
| public: | |
| vector<vector<int> > levelOrder(TreeNode *root) { | |
| // Start typing your C/C++ solution below | |
| // DO NOT write int main() function | |
| vector<int> sol; | |
| vector<vector<int> > res; | |
| if(!root) return res; | |
| queue<TreeNode *> q1, q2; | |
| sol.push_back(root->val); | |
| res.push_back(sol); | |
| sol.clear(); | |
| if(root->left) q1.push(root->left); | |
| if(root->right) q1.push(root->right); | |
| while(!q1.empty()||!q2.empty()) | |
| { | |
| if(!q1.empty()) | |
| { | |
| while(!q1.empty()) | |
| { | |
| TreeNode *tmp=q1.front(); | |
| q1.pop(); | |
| sol.push_back(tmp->val); | |
| if(tmp->left) q2.push(tmp->left); | |
| if(tmp->right) q2.push(tmp->right); | |
| } | |
| res.push_back(sol); | |
| sol.clear(); | |
| } | |
| if(!q2.empty()) | |
| { | |
| while(!q2.empty()) | |
| { | |
| TreeNode *tmp=q2.front(); | |
| q2.pop(); | |
| sol.push_back(tmp->val); | |
| if(tmp->left) q1.push(tmp->left); | |
| if(tmp->right) q1.push(tmp->right); | |
| } | |
| res.push_back(sol); | |
| sol.clear(); | |
| } | |
| } | |
| return res; | |
| } | |
| }; |
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| /** | |
| * Definition for binary tree | |
| * struct TreeNode { | |
| * int val; | |
| * TreeNode *left; | |
| * TreeNode *right; | |
| * TreeNode(int x) : val(x), left(NULL), right(NULL) {} | |
| * }; | |
| */ | |
| class Solution { | |
| public: | |
| vector<vector<int> > levelOrder(TreeNode *root) { | |
| // Start typing your C/C++ solution below | |
| // DO NOT write int main() function | |
| vector<int> tmp; | |
| vector<vector<int>> res; | |
| if(!root) return res; | |
| queue<TreeNode*> level; | |
| level.push(root); | |
| int count=0; | |
| while(!level.empty()) | |
| { | |
| count=level.size(); | |
| while(count) | |
| { | |
| TreeNode* cur=level.front(); | |
| tmp.push_back(cur->val); | |
| count--; | |
| level.pop(); | |
| if(cur->left) | |
| level.push(cur->left); | |
| if(cur->right) | |
| level.push(cur->right); | |
| } | |
| res.push_back(tmp); | |
| tmp.clear(); | |
| } | |
| return res; | |
| } | |
| }; |
I always thought of using two queues or vectors. Never thought of using only one queue with simply counters. Grateful for being inspired by your blog. Thanks.
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