Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:算法
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:this
1 / \ 2 2 \ \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.spa
给定一棵树,判断它是不是对称的。即树的左子树是不是其右子树的镜像。.net
使用递归进行求解,先判断左右子结点是否相等,不等就返回false,相等就将左子结点的左子树与右子结果的右子结点进行比较操做,同时将左子结点的左子树与右子结点的左子树进行比较,只有两个同时为真是才返回true,不然返回false。code
树结点类递归
public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } }
算法实现类get
public class Solution { public boolean isSymmetric(TreeNode root) { if (root == null) { return true; } else { return isSame(root.left, root.right); } } private boolean isSame(TreeNode left, TreeNode right) { if (left == null && right == null) { return true; } if (left != null && right == null || left == null && right != null){ return false; } else { return left.val == right.val && isSame(left.left, right.right) && isSame(left.right, right.left); } } }