标签：nbsp || tree 逻辑 iter 解题思路 center span OWIN
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.
public bool IsSymmetric(TreeNode root) { return IsMirror(root?.left, root?.right); } public bool IsMirror(TreeNode left, TreeNode right) { bool flag; if (left == null && right == null) { flag = true; } else if (left == null || right == null) { flag = false; } else { if (left.val == right.val) { flag = IsMirror(left.left, right.right) && IsMirror(left.right, right.left); } else { flag = false; } } return flag; }
标签：nbsp || tree 逻辑 iter 解题思路 center span OWIN
原文地址：https://www.cnblogs.com/chucklu/p/10551576.html
