Symmetric Tree

　　题目链接

　　题目要求：

　　Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

　　For example, this binary tree is symmetric:

    1
   /   2   2
 / \ / 3  4 4  3

　　But the following is not:

    1
   /   2   2
   \      3    3

　　Note:
　　Bonus points if you could solve it both recursively and iteratively.

　　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}".

　　本题暂用递归解法，具体程序（4ms）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     void updateLVec(TreeNode *left, vector<int>& lVec)
13     {
14         if(left)
15         {
16             lVec.push_back(left->val);
17             if(left->left || left->right)
18             {
19                 if(left->left)
20                     updateLVec(left->left, lVec);
21                 else
22                     lVec.push_back(INT_MIN);
23                 if(left->right)
24                     updateLVec(left->right, lVec);
25                 else
26                     lVec.push_back(INT_MIN);
27             }
28         }
29     }
30
31     void updateRVec(TreeNode *right, vector<int>& rVec)
32     {
33         if(right)
34         {
35             rVec.push_back(right->val);
36             if(right->left || right->right)
37             {
38                 if(right->right)
39                     updateRVec(right->right, rVec);
40                 else
41                     rVec.push_back(INT_MIN);
42                 if(right->left)
43                     updateRVec(right->left, rVec);
44                 else
45                     rVec.push_back(INT_MIN);
46             }
47         }
48     }
49
50     bool isSymmetric(TreeNode* root) {
51         if(!root || (!root->left && !root->right))
52             return true;
53
54         if((!root->left && root->right) || (root->left && !root->right))
55             return false;
56
57         vector<int> lVec, rVec;
58         updateLVec(root->left, lVec);
59         updateRVec(root->right, rVec);
60
61         int szLVec = lVec.size();
62         int szRVec = rVec.size();
63         if(szLVec != szRVec)
64             return false;
65
66         for(int i = 0; i < szLVec; i++)
67         {
68             if(lVec[i] != rVec[i])
69                 return false;
70         }
71
72         return true;
73     }
74 };

　　虽然最终解决了问题，但代码还是复杂了。别人的代码（4ms）真叫一个简单呐：

 1 bool isSymmetric(TreeNode *root)
 2 {
 3     if (!root) return true;
 4     return isSymmetric(root->left, root->right);
 5 }
 6 bool isSymmetric(TreeNode *lt, TreeNode *rt)
 7 {
 8     if (!lt && !rt) return true;
 9     if (lt && !rt || !lt && rt || lt->val != rt->val) return false;
10     return isSymmetric(lt->left, rt->right) && isSymmetric(lt->right, rt->left);
11 }

Same Tree

　　题目链接

　　题目要求：

　　Given two binary trees, write a function to check if they are equal or not.

　　Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

　　有了上一道题的基础，这道题就很简单了，具体程序（0ms）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     bool isSameTree(TreeNode* p, TreeNode* q) {
13         if(!p && !q)
14             return true;
15         return isSameTreeSub(p, q);
16     }
17
18     bool isSameTreeSub(TreeNode *lt, TreeNode *rt)
19     {
20         if(!lt && !rt)
21             return true;
22         if((!lt && rt) || (lt && !rt) || (lt->val != rt->val))
23             return false;
24         return isSameTree(lt->left, rt->left) && isSameTree(lt->right, rt->right);
25     }
26 };