leetcode

题目：Symmetric Tree

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

For example, this binary tree is symmetric:

But the following is not:

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

个人思路：

1、注意到根节点的左右子树是对称的，且对称方向是左右对称，那么选取根节点的任意一颗子树，这里我选取右子树

2、将右子树中的所有父节点的左右子节点互换，如果原来的树是一颗镜像树，那么右子树经过这么处理后便与左子树相同，因此，只需要将处理之后的右子树与原来的左子树对比是否相同，若相同，则原树是镜像树，否则不是镜像树

代码：

 1 #include <stddef.h>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 struct TreeNode
 7 {
 8     int val;
 9     TreeNode *left;
10     TreeNode *right;
11     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
12 };
13 
14 class Solution
15 {
16 public:
17     bool isSymmetric(TreeNode *root)
18     {
19         if (root == NULL || (root->left == NULL && root->right == NULL))
20         {
21             return true;
22         }
23         if ((root->left == NULL && root->right != NULL) || (root->right == NULL && root->left != NULL))
24         {
25             return false;
26         }
27         //根节点及其左右子节点均不为空
28         swapLeftRight(root->right);
29 
30         return isSameTree(root->left, root->right);
31     }
32 private:
33     void swapLeftRight(TreeNode *root)
34     {
35         if (root == NULL)
36         {
37             return;
38         }
39 
40         TreeNode *tmp = root->right;
41         root->right = root->left;
42         root->left = tmp;
43 
44         swapLeftRight(root->left);
45         swapLeftRight(root->right);
46     }
47     bool isSameTree(TreeNode *first, TreeNode *second)
48     {
49         if (first == NULL && second == NULL)
50         {
51             return true;
52         }
53         if (first != NULL && second != NULL)
54         {
55             return first->val == second->val && isSameTree(first->left, second->left) && isSameTree(first->right, second->right);
56         }
57 
58         return false;
59     }
60 };
61 
62 int main()
63 {
64     Solution s;
65     TreeNode *root = new TreeNode(1);
66     root->left = new TreeNode(2);
67     root->right = new TreeNode(2);
68     //root->left->left = new TreeNode(3);
69     root->left->right = new TreeNode(3);
70     //root->right->left = new TreeNode(4);
71     root->right->right = new TreeNode(3);
72 
73     cout << s.isSymmetric(root) << endl;
74 
75     system("pause");
76 
77     return 0;
78 }

View Code

上网搜了一些帖子，发现我的方法过于麻烦，可以处理地更简单些，链接：http://www.cnblogs.com/remlostime/archive/2012/11/15/2772230.html，这是递归的算法，思路是从根节点的左右子节点开始判断，判断左右子节点值，再判断左节点的左子树与右节点的右子树是否相同，以及左节点的右子树与右节点的左子树是否相同

代码：

  1 #include <stddef.h>
  2 #include <iostream>
  3 
  4 using namespace std;
  5 
  6 struct TreeNode
  7 {
  8     int val;
  9     TreeNode *left;
 10     TreeNode *right;
 11     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 12 };
 13 
 14 class Solution
 15 {
 16 public:
 17     bool isSymmetric(TreeNode *root)
 18     {
 19         /*
 20         if (root == NULL || (root->left == NULL && root->right == NULL))
 21         {
 22             return true;
 23         }
 24         if ((root->left == NULL && root->right != NULL) || (root->right == NULL && root->left != NULL))
 25         {
 26             return false;
 27         }
 28         //根节点及其左右子节点均不为空
 29         swapLeftRight(root->right);
 30 
 31         return isSameTree(root->left, root->right);
 32         */
 33 
 34         if (root == NULL)
 35         {
 36             return true;
 37         }
 38 
 39         return isSame(root->left, root->right);
 40     }
 41 private:
 42     bool isSame(TreeNode *left, TreeNode *right)
 43     {
 44         if (left == NULL && right == NULL)
 45         {
 46             return true;
 47         }
 48         if (left == NULL || right == NULL)
 49         {
 50             return false;
 51         }
 52 
 53         return left->val == right->val && isSame(left->left, right->right) && isSame(left->right, right->left);
 54     }
 55     void swapLeftRight(TreeNode *root)
 56     {
 57         if (root == NULL)
 58         {
 59             return;
 60         }
 61 
 62         TreeNode *tmp = root->right;
 63         root->right = root->left;
 64         root->left = tmp;
 65 
 66         swapLeftRight(root->left);
 67         swapLeftRight(root->right);
 68     }
 69     bool isSameTree(TreeNode *first, TreeNode *second)
 70     {
 71         if (first == NULL && second == NULL)
 72         {
 73             return true;
 74         }
 75         if (first != NULL && second != NULL)
 76         {
 77             return first->val == second->val && isSameTree(first->left, second->left) && isSameTree(first->right, second->right);
 78         }
 79 
 80         return false;
 81     }
 82 };
 83 
 84 int main()
 85 {
 86     Solution s;
 87     TreeNode *root = new TreeNode(1);
 88     root->left = new TreeNode(2);
 89     root->right = new TreeNode(2);
 90     //root->left->left = new TreeNode(3);
 91     root->left->right = new TreeNode(3);
 92     //root->right->left = new TreeNode(4);
 93     root->right->right = new TreeNode(3);
 94 
 95     cout << s.isSymmetric(root) << endl;
 96 
 97     system("pause");
 98 
 99     return 0;
100 }

View Code

非递归的算法，链接：http://blog.csdn.net/sunbaigui/article/details/8981333

思路：

1、类似层次遍历的方法，根节点的左右子树分别进行各自的层次遍历，且相同层次上的遍历顺序相反

2、左子树从左往右，右子树从右往左，如果原树是镜像树，那么遍历出来的节点以及节点值应该相同，否则不是镜像树

代码：

  1 #include <stddef.h>
  2 #include <iostream>
  3 #include <queue>
  4 
  5 using namespace std;
  6 
  7 struct TreeNode
  8 {
  9     int val;
 10     TreeNode *left;
 11     TreeNode *right;
 12     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 13 };
 14 
 15 class Solution
 16 {
 17 public:
 18     bool isSymmetric(TreeNode *root)
 19     {
 20         /*
 21         if (root == NULL || (root->left == NULL && root->right == NULL))
 22         {
 23             return true;
 24         }
 25         if ((root->left == NULL && root->right != NULL) || (root->right == NULL && root->left != NULL))
 26         {
 27             return false;
 28         }
 29         //根节点及其左右子节点均不为空
 30         swapLeftRight(root->right);
 31 
 32         return isSameTree(root->left, root->right);
 33         */
 34         /*
 35         if (root == NULL)
 36         {
 37             return true;
 38         }
 39 
 40         return isSame(root->left, root->right);
 41         */
 42 
 43         if (root == NULL)
 44         {
 45             return true;
 46         }
 47 
 48         queue<TreeNode *> left, right;
 49         left.push(root->left);
 50         right.push(root->right);
 51 
 52         while (!left.empty() && !right.empty())
 53         {
 54             TreeNode *leftNode = left.front();
 55             TreeNode *rightNode = right.front();
 56             left.pop();
 57             right.pop();
 58 
 59             if (leftNode == NULL && rightNode == NULL)
 60             {
 61                 continue;
 62             }
 63             if (leftNode == NULL || rightNode == NULL)
 64             {
 65                 return false;
 66             }
 67             if (leftNode->val != rightNode->val)
 68             {
 69                 return false;
 70             }
 71 
 72             left.push(leftNode->left);
 73             left.push(leftNode->right);
 74             right.push(rightNode->right);
 75             right.push(rightNode->left);
 76         }
 77 
 78         return true;
 79 
 80     }
 81 private:
 82     bool isSame(TreeNode *left, TreeNode *right)
 83     {
 84         if (left == NULL && right == NULL)
 85         {
 86             return true;
 87         }
 88         if (left == NULL || right == NULL)
 89         {
 90             return false;
 91         }
 92 
 93         return left->val == right->val && isSame(left->left, right->right) && isSame(left->right, right->left);
 94     }
 95     void swapLeftRight(TreeNode *root)
 96     {
 97         if (root == NULL)
 98         {
 99             return;
100         }
101 
102         TreeNode *tmp = root->right;
103         root->right = root->left;
104         root->left = tmp;
105 
106         swapLeftRight(root->left);
107         swapLeftRight(root->right);
108     }
109     bool isSameTree(TreeNode *first, TreeNode *second)
110     {
111         if (first == NULL && second == NULL)
112         {
113             return true;
114         }
115         if (first != NULL && second != NULL)
116         {
117             return first->val == second->val && isSameTree(first->left, second->left) && isSameTree(first->right, second->right);
118         }
119 
120         return false;
121     }
122 };
123 
124 int main()
125 {
126     Solution s;
127     TreeNode *root = new TreeNode(1);
128     root->left = new TreeNode(2);
129     root->right = new TreeNode(2);
130     //root->left->left = new TreeNode(3);
131     root->left->right = new TreeNode(3);
132     //root->right->left = new TreeNode(4);
133     root->right->right = new TreeNode(3);
134 
135     cout << s.isSymmetric(root) << endl;
136 
137     system("pause");
138 
139     return 0;
140 }

View Code

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原文地址：https://www.cnblogs.com/laihaiteng/p/3796413.html