R8-二叉树篇
思路：
DFS搜索+递归分解成子问题逐步判断即可。
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public boolean isSymmetric(TreeNode root) {return root==null || recur(root.left,root.right);}boolean recur(TreeNode L,TreeNode R){if (L==null && R==null) return true;if (L==null || R==null || L.val!=R.val) return false;return recur(L.left,R.right)&&recur(L.right,R.left);}
}
真-没落的java