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.
判断两个节点是否对称，只需保证，这两个点是否相等，这个点的左子节点和另一个点的有子节点是否相等，这个点的右子节点和另一个点的左子节点是否相等。
最初的两个点必须是根节点的左右子节点。
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 bool key;
13 bool isSymmetric(TreeNode *root) {
14 // Start typing your C/C++ solution below
15 // DO NOT write int main() function
16 key = true;
17 if(root == NULL)
18 return key;
19 cmp(root->left,root->right);
20 return key;
21 }
22 void cmp(TreeNode *node1,TreeNode *node2)
23 {
24 if(node1 == NULL && node2 == NULL)
25 return;
26 if(node1 == NULL || node2 == NULL)
27 {
28 key = false;
29 return;
30 }
31 if(node1->val != node2->val)
32 {
33 key = false;
34 return;
35 }
36 cmp(node1->left,node2->right);
37 if(key == false)
38 return;
39 cmp(node1->right,node2->left);
40 }
41 };
42
43 /**
44 * Definition for binary tree
45 * struct TreeNode {
46 * int val;
47 * TreeNode *left;
48 * TreeNode *right;
49 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
50 * };
51 */
52 class Solution {
53 public:
54 bool isSymmetric(TreeNode *root) {
55 // Start typing your C/C++ solution below
56 // DO NOT write int main() function
57 queue<TreeNode*> q1;
58 queue<TreeNode*> q2;
59 if(root == NULL)
60 return true;
61 q1.push(root->left);
62 q2.push(root->right);
63 while(!q1.empty())
64 {
65 TreeNode * tmp1 = q1.front();
66 TreeNode * tmp2 = q2.front();
67 q1.pop();
68 q2.pop();
69 if(tmp1 == NULL && tmp2 == NULL)
70 continue;
71 if(tmp1 == NULL || tmp2 == NULL)
72 return false;
73 if(tmp1->val != tmp2->val)
74 return false;
75 q1.push(tmp1->left);
76 q1.push(tmp1->right);
77 q2.push(tmp2->right);
78 q2.push(tmp2->left);
79 }
80 }
81
82 };
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