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
判定一棵树是不是对称的,和[url=http://kickcode.iteye.com/blog/2276006]Same Tree[/url]这道题思想一样,same tree是判断两个数是不是一样,这道题目是判断左子树是不是右子树的镜像,用递归解决。代码如下:
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
判定一棵树是不是对称的,和[url=http://kickcode.iteye.com/blog/2276006]Same Tree[/url]这道题思想一样,same tree是判断两个数是不是一样,这道题目是判断左子树是不是右子树的镜像,用递归解决。代码如下:
/**
* 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 isSym(root.left, root.right);
}
public boolean isSym(TreeNode leftNode, TreeNode rightNode) {
if(leftNode == null && rightNode == null) return true;
if(leftNode != null && rightNode != null) {
if(leftNode.val != rightNode.val) return false;
return isSym(leftNode.left, rightNode.right) && isSym(leftNode.right, rightNode.left);
}
return false;
}
}