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【solution】:
An easy method to solve this problem is recursive algorithm. The process is comparing right node and left node of every node. And call this method recursively.
Once right node != left node, return false. Otherwise, return true.
【code】:
public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root == null) return true;
return isSymmetric(root.left, root.right);
}
public static boolean isSymmetric(TreeNode left, TreeNode right) {
if(left == null && right == null) return true;
if(left == null || right == null) return false;
if(left.val == right.val) return isSymmetric(left.left, right.right)
&& isSymmetric(left.right, right.left);
return false;
}
}
192 / 192 test
cases passed. Runtime: 1
msfinally, welcome criticism!