226. Invert Binary Tree

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

Thoughts

• Seems like a pretty simple problem, I would imagine that doing this recursively would be the easiest way to go about it

  • Simply swap the left and right nodes, then recursively swap the child node's left and right nodes

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) {
        // Handle null case
        if (root == null) return null;

        // Recursively swap the left and right nodes
        if (root.left != null && root.right != null) {
            TreeNode temp = root.right;
            root.right = root.left;
            root.left = temp;
            invertTree(root.left);
            invertTree(root.right);
        } else if (root.left == null) {
            root.left = root.right;
            root.right = null;
            invertTree(root.left);
        } else {
            root.right = root.left;
            root.left = null;
            invertTree(root.right);
        }

        return root;
    }
}

Time Complexity: O(n)
Space Complexity: O(n)

Conclusion

Pretty simple algorithm to implement and understand, good refresher on tree stuff. My space complexity is O(n) and I feel like there is a more space efficient algorithm to complete the task. I will revisit this problem in the future and try to devise a better algorithm.