Recursive DFS: Swap Children at Every Node (Classic Tree Inversion)

Intuition

To invert a binary tree, we simply need to swap the left and right child of every node in the tree.

Since a binary tree is naturally defined recursively (each node is the root of its own subtree), we can apply the same operation to every node:

• Swap its left and right children
• Recursively invert the left subtree
• Recursively invert the right subtree

By doing this for every node, the entire tree gets inverted.

Approach

We solve the problem using Depth-First Search (DFS) recursion.

1. Base Case
   • If the current node is nullptr, return nullptr.
   • If the node is a leaf (both children are nullptr), no work is required.
2. Swap Children
   • Swap the left and right pointers of the current node.
3. Recursive Calls
   • Recursively invert the left subtree.
   • Recursively invert the right subtree.
4. Return Root
   • Return the root after processing the entire subtree.

Because recursion processes every node exactly once, the whole tree gets inverted efficiently.

Complexity

• Time complexity:

  \[O(n)\]

  Every node in the tree is visited exactly once.

• Space complexity:

  \[O(h)\]

  Where h is the height of the tree due to recursion stack.

  • Worst case (skewed tree): \(O(n)\)
  • Balanced tree: \(O(\log n)\)

Code

/**
 * 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) {
        if(root == nullptr || (!root->left && !root->right)) return root;
        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};