Invert Binary Tree

• LeetCode: Invert Binary Tree

• Problem:

Invert a binary tree.

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

to

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

• Trivial:

This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

• Analysis:

Invert every node recursively.

• Code - Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        invertTree(root.left);
        invertTree(root.right);
        TreeNode node = root.left;
        root.left = root.right;;
        root.right = node;
        return root;
    }
}

• Code - C++:

/**
 * 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 == NULL) {
            return NULL;
        }
        invertTree(root->left);
        invertTree(root->right);
        TreeNode* node = root->left;
        root->left = root->right;;
        root->right = node;
        return root;
    }
};

• Code - 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 invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if root == None:
            return None
        self.invertTree(root.left)
        self.invertTree(root.right)
        node = root.left
        root.left = root.right
        root.right = node
        return root

• Code - C:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* invertTree(struct TreeNode* root) {
    if (root == 0) {
            return 0;
        }
        invertTree(root->left);
        invertTree(root->right);
        struct TreeNode* node = root->left;
        root->left = root->right;;
        root->right = node;
        return root;
}

• Time Complexity: O(N), N is the number of nodes.
• Space Complexity: in-memory