Symmetric Tree（对称二叉树）

来源：https://leetcode.com/problems/symmetric-tree

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

But the following [1,2,2,null,3,null,3] is not:

递归：

1. 若根节点为null，返回true

2. 若根节点不为null，对其左右子树进行递归地比较：若两个节点都为null，返回true；若一个为null，另一个不为null，返回false；若两个节点（A,B）值相等，则返回 (A.left, B.right) 的比较结果 && (A.right, B.left) 的比较结果。

Java

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public boolean compareLR(TreeNode left, TreeNode right) {
12         if(left == null && right == null) {
13             return true;
14         }
15         if((left == null && right != null) || (left != null && right == null)) {
16             return false;
17         }
18         if(left.val == right.val) {
19             return compareLR(left.right, right.left) && compareLR(left.left, right.right);
20         }
21         return false;
22     }
23     public boolean isSymmetric(TreeNode root) {
24         if(root == null) {
25             return true;
26         }
27         return compareLR(root.left, root.right);
28     }
29 }

Python

 1 # -*- coding:utf-8 -*-
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 class Solution:
 8     def isSymmetrical(self, pRoot):
 9         def comRoot(left, right):
10             if not left:
11                 return not right
12             if not right or (left.val != right.val):
13                 return False
14             return comRoot(left.right, right.left) and comRoot(left.left, right.right)
15         if not pRoot:
16             return True
17         return comRoot(pRoot.left, pRoot.right)

迭代：

1. 若根节点为null，返回true

2. 若根节点不为null，将其左右两个子节点入队列

3. 从队列中取出两个节点，若两个节点都为null，返回true；若一个为null，另一个不为null，返回false；若两个节点（A,B）值相等，将他们的两个子节点按照A.left，B.right，A.right，B.left的顺序入队列，重复执行该步，直到队列中不再有节点。