标签：tree node lse follow solution self iss col null points
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:
1 / 2 2 / \ / 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 class Solution { 11 public boolean check(TreeNode l, TreeNode r) { 12 if (l == null && r == null) return true; 13 if (l == null && r != null) return false; 14 if (l != null && r == null) return false; 15 if (l.val != r.val) return false; 16 if (!check(l.left, r.right)) return false; 17 if (!check(l.right, r.left)) return false; 18 return true; 19 } 20 public boolean isSymmetric(TreeNode root) { 21 if (root == null) return true; 22 return check(root.left, root.right); 23 } 24 }
标签：tree node lse follow solution self iss col null points
原文地址：https://www.cnblogs.com/hyxsolitude/p/12324280.html
