Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:
1 / \ 2 2 \ \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
思路:本题就是判断一个二叉树是不是对称的,树的每一层都可以用水仙花数的判断方法进行判断,只要有一层不是水仙花数就不是对称的树。
代码如下(已通过leetcode)
public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root==null) return true;
return isequal(root.left,root.right);
}
//
public boolean isequal(TreeNode p,TreeNode q) {
if(p==null || q==null) return p==q;
return p.val==q.val && isequal(p.left,q.right) && isequal(p.right,q.left);
}
}