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.
判断一棵树是否是对称的。
/** * 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; if( root.left == null && root.right == null) return true; if( root.left == null || root.right == null) return false; return getResult(root.left,root.right); } public boolean getResult(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; if( getResult(left.left,right.right) ) return getResult(left.right,right.left); return false; } }
也可以使用队列来解决这个问题。