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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Recursive:

public class Solution {
 public boolean isSymmetric(TreeNode root) {
   if(root==null){
     return true;
   }
   return isSymmetric(root.left, root.right);
 }
 
 public boolean isSymmetric(TreeNode left, TreeNode right){
   if(left==null&&right==null){
     return true;
   }
   if((left==null&&right!=null)||(left!=null&right==null)){
     return false;
   }
   return left.val==right.val && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
 }
}

Non-Recursive/Iteratively:

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