题目描述：

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.

思路：

判断一棵树是否对称。bonus：使用递归和迭代两种方法实现

递归：判断根节点是否为空，空则是对称。否则判断左右子节点。

如果左右均为空，则对称。如果左空或右空，则不对称。

如果左子节点的值不等于右子节点，则不对称。

否则比较左子节点的左子节点和右子节点的右子节点，以及左子节点的右子节点和右子节点的左子节点。

依次递归

迭代：

用一个栈或者队列存储树种所有的值，模拟地规定的过程

/*public class TreeNode {

int val;

TreeNode left;

TreeNode right;

TreeNode(int x) { val = x; }

}*/

1 public class Solution101 { 2     public boolean isSymmetric(TreeNode root) { 3         if(root == null) return true; 4         return isSymmetric(root.left,root.right); 5     } 6     public boolean isSymmetric(TreeNode left,TreeNode right){ 7         if(left == null && right == null) return true; 8         if(left == null || right == null) return false; 9         if(left.val != right.val) return false;10         return isSymmetric(left.left, right.right) && isSymmetric(left.right,right.left);11     }12     public static void main(String[] args) {13         // TODO Auto-generated method stub14         Solution101 solution101 = new Solution101();15         TreeNode root = new TreeNode(1);16         root.left = new TreeNode(2);17         root.right = new TreeNode(2);18         root.left.left = new TreeNode(3);19         root.left.right = new TreeNode(4);20         root.right.left = new TreeNode(4);21         root.right.right = new TreeNode(3);22         if(solution101.isSymmetric(root) == true)23             System.out.println("1");24         else {25             System.out.println("0");26         }27     }28 29 }