# 226. Invert Binary Tree

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

# Solution

Approach 1: Recursion.

Approach 2: Iteration with BFS.

Approach 3: Iteration with DFS.

# Code (Python)

Approach 1:

    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return root
        self.invertTree(root.left)
        self.invertTree(root.right)
        root.left, root.right = root.right, root.left
        return root

Approach 3:

    def invertTree(self, root):
        if not root:
            return root
        stack = [root]
        while stack:
            node = stack.pop()
            node.left, node.right = node.right, node.left
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        return root

# 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:
    TreeNode* invertTree(TreeNode* root) {
        if (root)
        {
            TreeNode *left = invertTree(root->left);
            TreeNode *right = invertTree(root->right);
            root->left = right;
            root->right = left;
        }
        return root;
    }
};

Approach 2:

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        queue<TreeNode*> q;
        if (root) q.push(root);
        while (!q.empty())
        {
            TreeNode *node = q.front();
            q.pop();
            std::swap(node->left, node->right);
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
        return root;
    }
};