101. Symmetric Tree (判斷二叉樹是否對稱)
阿新 • • 發佈:2019-01-31
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.
/** * 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 judge(root.left, root.right); } public boolean judge(TreeNode left,TreeNode right){ if(left==null&&right==null) return true; if(left==null||right==null) return false; if(left.val!=right.val) return false; return judge(left.left, right.right)&&judge(left.right, right.left); } }