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.
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1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public boolean isSymmetric(TreeNode root) { 12 if(root == null) return true; 13 return judge(root.left,root.right); 14 } 15 public boolean judge(TreeNode left, TreeNode right){ 16 if(left == null && right == null) return true; 17 if((left == null && right != null) || (left != null && right == null)) return false; 18 return judge(left.left,right.right) && judge(left.right,right.left) && left.val == right.val; 19 } 20 }