【题解】【BT】【Leetcode】Binary Tree Level Order Traversal

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：

重点在于每层元素个数不定，如何标记一层的结束，往堆栈里push很多NULL来表示空位这种方案，会造成Memory Limit Exceeded。

可以采取记下每层的NULL数量，下层翻倍这种方式计数，满额push标记NULL作为一层的结束

代码：

 vector<vector<int> > levelOrder(TreeNode *root) {
     vector<vector<int> > orders;
     if(root == NULL)
         return orders;

     vector<int> vtmp;
     queue<TreeNode*> tque;
     tque.push(root);
     tque.push(NULL);

     int size = ;
     int count = ;
     int zero = ;//该层的NULL数
     while(!tque.empty()){
         TreeNode * tmp = tque.front();//$$$$
         tque.pop();//void pop()

         if(tmp == NULL){//NULL标识一层的结束
             if(!vtmp.empty())//最后一行可能不满
                 orders.push_back(vtmp);
             vtmp.clear();
             continue;
         }
         vtmp.push_back(tmp->val);
         if(tmp->left != NULL){
             tque.push(tmp->left);
             count++;
         }
         else
             zero++;
         if(tmp->right != NULL){
             tque.push(tmp->right);
             count++;
         }
         else
             zero++;

         if(count + zero == size){
             tque.push(NULL);
             count = ;
             size *= ;
             zero = zero * ;
         }
     }
     return orders;
 }