leetcode--101. Symmetric Tree

1、问题描述

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

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

2、边界条件：root==null

3、思路：1）递归，对称结构是镜像；左手和右手的关系。所以应该左子树和右子树比较，右子树和左子树比较是否相同same。但是首先，从root开始分为左右子树。

base case：左右都为空--true，左右有一个为空--false

2）迭代：一层一层的比较，可以利用队列，前后取值比较是否相同。

4、代码实现

1）递归

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

    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        }
        if (p == null || q == null) {
            return false;
        }

        return p.val == q.val
               && isSameTree(p.left, q.right)
               && isSameTree(p.right, q.left);
    }
}

2）迭代

5、api