Symmetric Tree - Leetcode

/**
 * 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 isSymmetricHelper(root.left, root.right);   
    }
    private boolean isSymmetricHelper(TreeNode left, TreeNode right){
        if(right == null && left == null)
           return true;
        if(right == null || left == null)
           return false;
        return right.val == left.val && isSymmetricHelper(left.left, right.right) && isSymmetricHelper(left.right, right.left);   
    }
}

迭代版本：

/**
 * 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;
        Stack<TreeNode> s = new Stack<>();
        s.push(root.left);
        s.push(root.right);
        
        while(!s.empty()){
            TreeNode nd1 = s.pop();
            TreeNode nd2 = s.pop();
            if(nd1==null && nd2==null)
               continue;
            if(nd1==null || nd2==null)
               return false;
            if(nd1.val!=nd2.val)
               return false;
            
            s.push(nd1.left); s.push(nd2.right);
            s.push(nd1.right); s.push(nd2.left);
        }
        return true;
    }
}

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.