binary-tree-level-order-traversal I、II——输出二叉树的数字序列

I

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.

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
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

 /**
  * 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) {
         vector<vector<int>> res;
         if(root==NULL) return res;
         queue<TreeNode*> q;
         q.push(root);
         while(!q.empty()){
             int n=q.size();
             vector<int> v;
             for(int i=;i<n;i++){
                 TreeNode *cur=q.front();
                 q.pop();
                 v.push_back(cur->val);
                 if(cur->left!=NULL)
                     q.push(cur->left);
                 if(cur->right!=NULL)
                     q.push(cur->right);
             }
             res.push_back(v);
         }
         return res;
     }
 };

II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree{3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

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

先将结果v存入stack中，最后在从stack倒入res形成倒序，未找到其他好的方法

 /**
  * 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> > levelOrderBottom(TreeNode *root) {
         vector<vector<int>> res;
         if(root==NULL) return res;
         stack<vector<int>> s;
         queue<TreeNode*> q;
         q.push(root);
         while(!q.empty()){
             int n=q.size();
             vector<int> v;
             for(int i=;i<n;i++){
                 TreeNode *cur=q.front();
                 q.pop();
                 v.push_back(cur->val);
                 if(cur->left!=NULL)
                     q.push(cur->left);
                 if(cur->right!=NULL)
                     q.push(cur->right);
 
             }
             s.push(v);
         }
         while(!s.empty()){
             res.push_back(s.top());
             s.pop();
         }
         return res;
     }
 };