LeetCode: Symmetric Tree 解题报告

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:

    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.

SOL 1 & 2:

两个解法，递归和非递归.

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     // solution 1:
12     public boolean isSymmetric(TreeNode root) {
13         if (root == null) {
14             return true;
15         }
16         
17         return isSymmetricTree(root.left, root.right);
18     }
19     
20     /*
21      * 判断两个树是否互相镜像 
22         (1) 根必须同时为空，或是同时不为空
23      * 
24      * 如果根不为空: 
25         (1).根的值一样 
26         (2).r1的左树是r2的右树的镜像 
27         (3).r1的右树是r2的左树的镜像
28      */
29     public boolean isSymmetricTree1(TreeNode root1, TreeNode root2) {
30         if (root1 == null && root2 == null) {
31             return true;
32         }
33         
34         if (root1 == null || root2 == null) {
35             return false;
36         }
37         
38         return root1.val == root2.val && 
39              isSymmetricTree(root1.left, root2.right) && 
40              isSymmetricTree(root1.right, root2.left);
41     }
42     
43     // solution 2:
44     public boolean isSymmetricTree(TreeNode root1, TreeNode root2) {
45         if (root1 == null && root2 == null) {
46             return true;
47         }
48         
49         if (root1 == null || root2 == null) {
50             return false;
51         }
52         
53         Stack<TreeNode> s1 = new Stack<TreeNode>();
54         Stack<TreeNode> s2 = new Stack<TreeNode>();
55         
56         // 一定记得初始化
57         s1.push(root1);
58         s2.push(root2);
59         
60         while (!s1.isEmpty() && !s2.isEmpty()) {
61             TreeNode cur1 = s1.pop();
62             TreeNode cur2 = s2.pop();
63             
64             if (cur1.val != cur2.val) {
65                 return false;
66             }
67             
68             if (cur1.left != null && cur2.right != null) {
69                 s1.push(cur1.left);
70                 s2.push(cur2.right);
71             } else if (!(cur1.left == null && cur2.right == null)) {
72                 return false;
73             }
74             
75             if (cur1.right != null && cur2.left != null) {
76                 s1.push(cur1.right);
77                 s2.push(cur2.left);
78             } else if (!(cur1.right == null && cur2.left == null)) {
79                 return false;
80             }
81         }
82         
83         return true;
84     }
85 }

View Code

2015.1.20:

简化了递归的解法，root与root本身是一对对称树。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSymmetric(TreeNode root) {
12         return isMirror(root, root);
13     }
14     
15     public boolean isMirror(TreeNode r1, TreeNode r2) {
16         if (r1 == null && r2 == null) {
17             return true;
18         } else if (r1 == null || r2 == null) {
19             return false;
20         }
21         
22         return r1.val == r2.val
23              && isMirror(r1.left, r2.right)
24              && isMirror(r1.right, r2.left);
25     }
26 }

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsSymmetric.java