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:
对称二叉树
But the following [1,2,2,null,3,null,3] is not:
非对称二叉树
Note:Bonus points if you could solve it both recursively and iteratively.
给定一个二叉树，检查是否是关于自己的镜像。
算法分析
方法一（递归）
- 分析
一个树如果是另一个树的镜像，需要满足如下条件：- 两棵树的根相等
- 左子树是右子树的镜像
- Java代码
/**
* 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) {
return isMirror(root, root);//采用递归的思想
}
public boolean isMirror(TreeNode root1, TreeNode root2) {
if (root1 == null && root2 == null) return true;
if (root1 == null || root2 == null) return false;
//节点相同，左子树与右子树为镜像，然后递归
return (root1.val == root2.val) && isMirror(root1.left, root2.right) && isMirror(root1.right, root2.left);
}
}
