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.

題目

給定二叉樹，判斷其是否左右對稱

思路

DFS

程式碼

 1 /* 
 2 Time:O(n)，
 3 Space: O(logn) 即 O(h) 樹的deepest hight
 4 */
 5 public class Solution {
 6     public boolean isSymmetric(TreeNode root) {
 7         if
 
 (root == null) return true;
 8         return isSymmetric(root.left, root.right);
 9     }
10     private static boolean isSymmetric(TreeNode p, TreeNode q) {
11         if (p == null && q == null) return true;   // 終止條件
12         if (p == null || q == null) return false;  // 終止條件
13         return
 
 p.val == q.val      // 三方合併
14                 && isSymmetric(p.left, q.right)
15                 && isSymmetric(p.right, q.left);
16     }
17 }