[Leetcode] Symmetric Tree

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.

Solution 1:　　 recursive

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        return mySymmetric(root.left,root.right);
        
    }

    private boolean mySymmetric(TreeNode left, TreeNode right) {
        // TODO Auto-generated method stub
        if(left==null&&right==null)
            return true;
        if(left==null||right==null)
            return false;
        return (left.val==right.val)&&mySymmetric(left.right, right.left)&&mySymmetric(left.left, right.right);
    }
}

Solution 2:　　iterative

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        //iterative way
        if(root==null)
            return true;
        Stack<TreeNode> sl=new Stack<TreeNode>();
        Stack<TreeNode> sr=new Stack<TreeNode>();
        sl.add(root.left);
        sr.add(root.right);
        while(!sl.isEmpty()&&!sr.isEmpty()){
            TreeNode n1=sl.pop();
            TreeNode n2=sr.pop();
            if(n1==null&&n2==null)
                continue;
            if(n1==null||n2==null)
                return false;
            if(n1.val!=n2.val)
                return false;
            sl.push(n1.left);
            sl.push(n1.right);
            sr.push(n2.right);
            sr.push(n2.left);
        }
        return true;
    }
}