leetcode 101. Symmetric Tree

题目内容

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

For example, this binary tree is symmetric:

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

But the following is not:

1
/ \
2   2
\   \
3    3

Note:

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

题目分析

这道题是上一道题 leetcode 100. Same Tree 的改良版。

思路基本一样还是要通过不断的递归，与对应的结点比较。这里稍微特殊的地方在于，我们假设，当树的深度>=2的时候，我们可以把一棵树分成两棵树来看待。

当我们把问题回归为两棵树的时候，问题就简单多了，和上一道题一样，只要递归遍历对应结点就可以了。注意结点，一个不断往右走，一个不断往左走。

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

}
public boolean isSame(TreeNode left,TreeNode right)
{
if(left==null&&right==null) return true;
if(left==null||right==null) return false;

if(left.val!=right.val) return false;
else
return isSame(left.left,right.right)&&isSame(left.right,right.left);
}
}