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 [3,9,20,null,null,15,7]
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
二叉树的层次遍历使用队列来实现，很大程度上类似于求树的最大深度。
第一种解法：开始时当前结点数curLevelCount=1,nextLevelCount=0,根据curLevelCount来输入每一层的结点值，当其左右子树存在时，将左右子树存入队列，并nextLevelCount++，每弹出一个当前层的结点时，curLevelCount--，当其为0时，再创建一个新的vector。
C++：
1 vector<vector<int>> levelOrder(TreeNode* root) { 2 vector<vector<int>> res={}; 3 queue<TreeNode*> q; 4 if(!root) 5 return res; 6 q.push(root); 7 int curLevelCount=1,nextLevelCount=0; 8 vector<int> temp={}; 9 while(!q.empty()){ 10 TreeNode* cur=q.front(); 11 temp.push_back(cur->val); 12 q.pop(); 13 curLevelCount--; 14 if(cur->left){ 15 q.push(cur->left); 16 nextLevelCount++; 17 } 18 if(cur->right){ 19 q.push(cur->right); 20 nextLevelCount++; 21 } 22 if(curLevelCount==0){ 23 res.push_back(temp); 24 temp={}; 25 curLevelCount=nextLevelCount; 26 nextLevelCount=0; 27 } 28 } 29 return res; 30 }
第二种方法不用参数计算每一层结点个数，利用for循环，每一层时利用队列的大小来计算每一层的结点数。两层循环嵌套，效率比较低。
1 vector<vector<int>> levelOrder(TreeNode* root) { 2 vector<vector<int>> res={}; 3 queue<TreeNode*> q; 4 if(!root) 5 return res; 6 q.push(root); 7 while(!q.empty()){ 8 vector<int> temp={}; 9 for(int i=q.size();i>0;i--){ 10 TreeNode* cur=q.front(); 11 temp.push_back(cur->val); 12 q.pop(); 13 if(cur->left) 14 q.push(cur->left); 15 if(cur->right) 16 q.push(cur->right); 17 } 18 res.push_back(temp); 19 } 20 return res; 21 }
