226. Invert Binary Tree

Problem Description

Given the root of a binary tree, invert the tree, and return its root.

Example 1:

Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]

Example 2:

Input: root = [2,1,3]
Output: [2,3,1]

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 100]
• -100 <= Node.val <= 100

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        # Base case: if root is None, return None
        if not root:
            return None
            
        # Swap the left and right children
        root.left, root.right = root.right, root.left
        
        # Recursively invert the left and right subtrees
        self.invertTree(root.left)
        self.invertTree(root.right)
        
        return root

Java Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        // Base case: if root is null, return null
        if (root == null) {
            return null;
        }
        
        // Swap the left and right children
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        
        // Recursively invert the left and right subtrees
        invertTree(root.left);
        invertTree(root.right);
        
        return root;
    }
}

C++ Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        // Base case: if root is nullptr, return nullptr
        if (!root) {
            return nullptr;
        }
        
        // Swap the left and right children
        TreeNode* temp = root->left;
        root->left = root->right;
        root->right = temp;
        
        // Recursively invert the left and right subtrees
        invertTree(root->left);
        invertTree(root->right);
        
        return root;
    }
};

JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
    // Base case: if root is null, return null
    if (!root) {
        return null;
    }
    
    // Swap the left and right children
    const temp = root.left;
    root.left = root.right;
    root.right = temp;
    
    // Recursively invert the left and right subtrees
    invertTree(root.left);
    invertTree(root.right);
    
    return root;
};

C# Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public TreeNode InvertTree(TreeNode root) {
        // Base case: if root is null, return null
        if (root == null) {
            return null;
        }
        
        // Swap the left and right children
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        
        // Recursively invert the left and right subtrees
        InvertTree(root.left);
        InvertTree(root.right);
        
        return root;
    }
}