# 101. 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.

# Solution

Approach 1: Recursion.

Approach 2: Iteration. Roughly level order traversal (using a queue), but make sure 2 symmetric nodes go together so we can pop 2 nodes at once.

# Code (Python)

Approach 1:

    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        # recursive
        if not root:
            return True
        return self._is_symmetric(root.left, root.right)
    
    def _is_symmetric(self, tree1, tree2):
        if not tree1 and not tree2:
            return True
        if (tree1 or tree2) and not (tree1 and tree2):
            return False
        if tree1.val != tree2.val:
            return False
        return self._is_symmetric(tree1.left, tree2.right) and self._is_symmetric(tree1.right, tree2.left)

Approach 2:

    def isSymmetric(self, root):
        if not root:
            return True
        queue = deque([root.left, root.right])
        while queue:
            node1, node2 = queue.popleft(), queue.popleft()
            if node1 and node2:
                if node1.val != node2.val:
                    return False
                else:
                    queue.append(node1.left)
                    queue.append(node2.right)
                    queue.append(node1.right)
                    queue.append(node2.left)
            elif node1 or node2:
                return False
        return True

# Code (C++)

Approach 1:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (root == NULL) return true;
        return isMirror(root->left, root->right);
    }
    
    bool isMirror(TreeNode *t1, TreeNode *t2) {
        return (!t1 && !t2) ||
            (t1 && t2 &&
             t1->val == t2->val &&
             isMirror(t1->left, t2->right) &&
             isMirror(t1->right, t2->left));
    }
};

Approach 2:

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (root == NULL) return true;
        queue<TreeNode*> nodeQueue;
        nodeQueue.push(root->left);
        nodeQueue.push(root->right);
        while (!nodeQueue.empty())
        {
            TreeNode *left = nodeQueue.front();
            nodeQueue.pop();
            TreeNode *right = nodeQueue.front();
            nodeQueue.pop();
            if (left == NULL && right == NULL) continue;
            if (left == NULL || right == NULL) return false;
            if (left->val != right->val) return false;
            nodeQueue.push(left->left);
            nodeQueue.push(right->right);
            nodeQueue.push(left->right);
            nodeQueue.push(right->left);
        }
        return true;
    }
};