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.
题目
给定二叉树,判断其是否左右对称
思路
DFS
代码
/*
Time:O(n),
Space: O(logn) 即 O(h) 树的deepest hight
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return isSymmetric(root.left, root.right);
}
private static boolean isSymmetric(TreeNode p, TreeNode q) {
if (p == null && q == null) return true; // 终止条件
if (p == null || q == null) return false; // 终止条件
return p.val == q.val // 三方合并
&& isSymmetric(p.left, q.right)
&& isSymmetric(p.right, q.left);
}
}