Binary Tree Level Order Traversal

　　题目链接

　　题目要求：

　　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]
]

　　这道题利用宽度优先搜索就可以了，具体程序（8ms）如下：

 1 /**
 2  * Definition for a binary tree node.
 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>> retVec;
14         if(!root)
15             return retVec;
16
17         retVec.push_back(vector<int>{root->val});
18         queue<TreeNode *> que;
19         que.push(root);
20         while(true)
21         {
22             vector<int> vec;
23             queue<TreeNode *> q;
24             while(!que.empty())
25             {
26                 TreeNode *tree = que.front();
27                 que.pop();
28                 if(tree->left)
29                 {
30                     vec.push_back((tree->left)->val);
31                     q.push(tree->left);
32                 }
33                 if(tree->right)
34                 {
35                     vec.push_back((tree->right)->val);
36                     q.push(tree->right);
37                 }
38             }
39
40             if(!q.empty())
41             {
42                 que = q;
43                 retVec.push_back(vec);
44             }
45             else
46                 break;
47         }
48
49         return retVec;
50     }
51 };

Binary Tree Level Order Traversal II

　　题目链接

　　题目要求：

　　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]
]

　　这道题基本更上边的题目一样，我们只需要将上题的结果retVec最后再反转一下就可以了，即添加如下一行即可：

1 reverse(retVec.begin(), retVec.end());

　　这样的程序只要8ms，但下边基本一样的程序却要64ms：

 1 /**
 2  * Definition for a binary tree node.
 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>> retVec;
14         if(!root)
15             return retVec;
16
17         retVec.insert(retVec.begin(), vector<int>{root->val});
18         queue<TreeNode *> que;
19         que.push(root);
20         while(true)
21         {
22             vector<int> vec;
23             queue<TreeNode *> q;
24             while(!que.empty())
25             {
26                 TreeNode *tree = que.front();
27                 que.pop();
28                 if(tree->left)
29                 {
30                     vec.push_back((tree->left)->val);
31                     q.push(tree->left);
32                 }
33                 if(tree->right)
34                 {
35                     vec.push_back((tree->right)->val);
36                     q.push(tree->right);
37                 }
38             }
39
40             if(!q.empty())
41             {
42                 que = q;
43                 retVec.insert(retVec.begin(), vec);
44             }
45             else
46                 break;
47         }
48
49         return retVec;
50     }
51 };