Symmetric Tree

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.

Approach #1: C++.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
    }
}

　　

Approach #2: Java.[Using stack]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        TreeNode left, right;
        if (root.left != null) {
            if (root.right == null) 
                return false;
            stack.push(root.left);
            stack.push(root.right);
        } else if (root.right != null) 
            return false;
        
        while (!stack.empty()) {
            if (stack.size() % 2 != 0) 
                return false;
            right = stack.pop();
            left = stack.pop();
            if (right.val != left.val) 
                return false;
            
            if (left.left != null) {
                if (right.right == null) 
                    return false;
                stack.push(left.left);
                stack.push(right.right);
            } else if (right.right != null) 
                return false;
            
            if (left.right != null) {
                if (right.left == null) 
                    return false;
                stack.push(left.right);
                stack.push(right.left);
            } else if (right.left != null)
                return false;
            
        }
        return true;
    }
}

　　

Approach #3: Python.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root is None:
            return True
        else:
            return self.isMirror(root.left, root.right)
        
    def isMirror(self, left, right):
        if left is None and right is None:
            return True
        if left is None or right is None:
            return False
        
        if left.val == right.val:
            outPair = self.isMirror(left.left, right.right)
            inPair = self.isMirror(left.right, right.left)
            return outPair and inPair
        else:
            return False