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}".

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> > levelOrder(TreeNode *root) {
13         vector<vector<int>> res;
14         if(root==NULL) return res;
15         queue<TreeNode*> q;
16         q.push(root);
17         while(!q.empty()){
18             int n=q.size();
19             vector<int> v;
20             for(int i=0;i<n;i++){
21                 TreeNode *cur=q.front();
22                 q.pop();
23                 v.push_back(cur->val);
24                 if(cur->left!=NULL)
25                     q.push(cur->left);
26                 if(cur->right!=NULL)
27                     q.push(cur->right);
28             }
29             res.push_back(v);
30         }
31         return res;
32     }
33 };

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

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> > levelOrderBottom(TreeNode *root) {
13         vector<vector<int>> res;
14         if(root==NULL) return res;
15         stack<vector<int>> s;
16         queue<TreeNode*> q;
17         q.push(root);
18         while(!q.empty()){
19             int n=q.size();
20             vector<int> v;
21             for(int i=0;i<n;i++){
22                 TreeNode *cur=q.front();
23                 q.pop();
24                 v.push_back(cur->val);
25                 if(cur->left!=NULL)
26                     q.push(cur->left);
27                 if(cur->right!=NULL)
28                     q.push(cur->right);
29
30             }
31             s.push(v);
32         }
33         while(!s.empty()){
34             res.push_back(s.top());
35             s.pop();
36         }
37         return res;
38     }
39 };