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:

    1
   / \
  2   2
 / \ / \
3  4 4  3

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

    1
   / \
  2   2
   \   \
   3    3

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

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 class Solution {
     public boolean compareLR(TreeNode left, TreeNode right) {
         if(left == null && right == null) {
             return true;
         }
         if((left == null && right != null) || (left != null && right == null)) {
             return false;
         }
         if(left.val == right.val) {
             return compareLR(left.right, right.left) && compareLR(left.left, right.right);
         }
         return false;
     }
     public boolean isSymmetric(TreeNode root) {
         if(root == null) {
             return true;
         }
         return compareLR(root.left, root.right);
     }
 }

Python

 # -*- coding:utf-8 -*-
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 class Solution:
     def isSymmetrical(self, pRoot):
         def comRoot(left, right):
             if not left:
                 return not right
             if not right or (left.val != right.val):
                 return False
             return comRoot(left.right, right.left) and comRoot(left.left, right.right)
         if not pRoot:
             return True
         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。