LeetCode题解12：对称树

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

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

题目要求

1. 给定一棵树，判断这棵树是不是对称的，如果是则返回true，如果不是则返回false。

解题思路

采用递归法解，因为我们要判断这一颗树是不是对称的，我们可以对左边和右边的节点进行同时的对称遍历，判断是不是对称的，代码如下：

时间复杂度O(n) 超过100%的java提交

/**
 * 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) {
        // 采用递归方法遍历树，对左右两边进行镜像处理
        if(root == null) return true;
        TreeNode l=root.left,r=root.right;
        return recursive(l,r);
    }
    private boolean recursive(TreeNode l,TreeNode r){
        if(l==null&&r==null) return true;
        if(l==null||r==null) return false;
        if(l.val != r.val) return false;
        return recursive(l.left,r.right)&&recursive(l.right,r.left);
    }
}