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.
題意：判斷這顆樹是否對(duì)稱
思路：遞歸遍歷，判斷是否值相同。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public static boolean isSymmetric(TreeNode root) {
if(root == null) return true;
return isSame(root.left, root.right);
}
private static 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);
}
}
