AcWing 39. 对称的二叉树-Java
原题链接
简单
作者:
Fant.J
,
2019-05-14 11:39:35
,
所有人可见
,
阅读 967
/**
* 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) {
if(root == null){return true;}
return judge(root.left,root.right);
}
boolean judge(TreeNode left,TreeNode right){
// 左子树和右子树都是空
if(left == null && right == null){
return true;
}
// 左子树空,右子树非空
if(left == null && right != null){
return false;
}
// 右子树空,左子树非空
if(right == null && left != null){
return false;
}
// 左右字子树不相等
if(left.val != right.val){
return false;
}
// 左右子树值相等,则判断其下层节点
if(left.val == right.val){
return judge(left.left,right.right)&&judge(left.right,right.left);
}
return true;
}
}