Solution #226 - hanzel-sc - 25/07/25 commit #1

Author: hanzel-sc

Trees (Binary Trees)/#226-Invert-Binary-Tree-Easy/Explanation.md

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+# Invert Binary Tree
+
+**Difficulty**: Easy  
+**Category**: Trees, Recursion, Binary Tree  
+**Leetcode Link**: [https://leetcode.com/problems/invert-binary-tree/](https://leetcode.com/problems/invert-binary-tree/)
+
+---
+
+## 📝 Introduction
+
+The **Invert Binary Tree** problem asks us to take a binary tree and **invert** it. 
+
+Given the `root` of a binary tree, our goal is to return the **root of the inverted tree**.
+
+**Expected Output**: Return the root of the new inverted tree (mirror image of the original).
+
+---
+
+## 💡 Approach & Key Insights
+
+The key insight is that at each node in the binary tree, we need to **swap** the left and right child. We can do this using:
+- **Recursion** (Depth-First Traversal)
+- **Iteration** using a Queue (Breadth-First Traversal)
+
+It's essentially a tree traversal problem and the recursive solution felt like the intuitive approach.
+
+---
+
+## 🛠️ Breakdown of Approaches
+
+### 1️⃣ Brute Force / Naive Approach
+
+#### Explanation:
+- Traverse the tree using any traversal (preorder/inorder/postorder).
+- For each node, store left and right subtrees separately.
+- Reconstruct the tree manually by reassigning children in inverted order.
+- This is impractical and complex for a simple task.
+
+#### Time Complexity:
+`O(n)` – We still visit every node once.
+
+#### Space Complexity:
+`O(n)` – We use space to store and reconstruct the tree.
+
+#### Example/Dry Run:
+
+> ⚠️ Skipping detailed dry run as this approach is rarely used in practice for this problem.
+
+---
+
+### 2️⃣ Optimized Approach (Recursive DFS)
+
+#### Explanation:
+- If the root is `NULL`, return `NULL`.
+- Recursively invert the left and right subtrees.
+- Swap the left and right pointers at the current node.
+- Return the node.
+
+```c
+struct TreeNode* invertTree(struct TreeNode* root) {
+    if (root == NULL)
+        return NULL;
+
+    struct TreeNode* temp = root->right;
+    root->right = root->left;
+    root->left = temp;
+
+    invertTree(root->right);
+    invertTree(root->left);
+
+    return root;
+}
+```
+
+#### Time Complexity:
+`O(n)` – Every node is visited exactly once.
+
+#### Space Complexity:
+`O(h)` – Due to the recursion stack. *h* is the height of the tree (O(log n) for balanced, O(n) for skewed).
+
+#### Example/Dry Run:
+
+Example Input (Tree):
+```
+    1
+   / \
+  2   3
+ / \
+4   5
+```
+
+Step-by-step:
+- At node 1: swap 2 and 3.
+- Recurse into 3 and 2.
+- At node 2: swap 4 and 5.
+- Final inverted tree:
+```
+    1
+   / \
+  3   2
+     / \
+    5   4
+```
+
+---
+
+### 3️⃣ Best / Final Optimized Approach (Iterative BFS)
+
+#### Explanation:
+- Use a queue to traverse the tree level by level.
+- At each node, swap the left and right children.
+- Enqueue non-null children for further processing.
+
+```c
+struct TreeNode* invertTree(struct TreeNode* root) {
+    if (root == NULL)
+        return NULL;
+
+    struct TreeNode* queue[100];
+    int front = 0, rear = 0;
+    queue[rear++] = root;
+
+    while (front < rear) {
+        struct TreeNode* node = queue[front++];
+
+        struct TreeNode* temp = node->left;
+        node->left = node->right;
+        node->right = temp;
+
+        if (node->left) queue[rear++] = node->left;
+        if (node->right) queue[rear++] = node->right;
+    }
+
+    return root;
+}
+```
+
+#### Time Complexity:
+`O(n)` – Each node is visited once.
+
+#### Space Complexity:
+`O(n)` – In the worst case, all nodes may be in the queue (for a full binary tree).
+
+#### Example/Dry Run:
+
+Input Tree:
+```
+    1
+   / \
+  2   3
+```
+
+- Queue: [1]
+- Process 1 → Swap 2 & 3
+- Queue: [3, 2]
+- Process 3 → Swap NULLs
+- Process 2 → Swap NULLs
+
+Final Tree:
+```
+    1
+   / \
+  3   2
+```
+
+---
+
+## 📊 Complexity Analysis
+
+| Approach         | Time Complexity | Space Complexity |
+|------------------|------------------|-------------------|
+| Brute Force      | O(n)             | O(n)              |
+| Optimized (DFS)  | O(n)             | O(h)              |
+| Best (BFS)       | O(n)             | O(n)              |
+
+---
+
+## 📉 Optimization Ideas
+
+- For deep trees where recursion depth could cause stack overflow, use **iterative BFS**.
+- If modifying the original tree is not allowed, we could clone and then invert, but that increases space usage.
+
+---
+
+## 📌 Example Walkthroughs & Dry Runs
+
+### Input Tree:
+```
+    4
+   / \
+  2   7
+ / \/ \
+1  3 6  9
+```
+
+### Process:
+- Swap 2 & 7 → Recurse
+- Swap 1 & 3 in subtree of 2
+- Swap 6 & 9 in subtree of 7
+
+### Output Tree:
+```
+    4
+   / \
+  7   2
+ / \/ \
+9  6 3  1
+```
+
+---
+
+## 🔗 Additional Resources
+
+- [YouTube: NeetCode - Invert Binary Tree](https://www.youtube.com/watch?v=OnSn2XEQ4MY)
+- [Binary Tree Traversal Explanation](https://www.geeksforgeeks.org/tree-traversals-inorder-preorder-and-postorder/)
+
+---
+
+**Author**: hanzel-sc  
+**Date**: 25/07/2025

Trees (Binary Trees)/#226-Invert-Binary-Tree-Easy/Invert Binary Tree.c

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+
+struct TreeNode* invertTree(struct TreeNode* root) {
+
+    if (root == NULL)
+    return NULL;
+
+    //A simple swap of the left and right nodes of the given tree
+    struct TreeNode *temp;
+
+    temp = root->right;
+    root->right = root->left;
+    root->left = temp;  
+
+    //recursive swapping at each node
+    invertTree(root->right);
+    invertTree(root->left);
+
+    return root;
+    
+}
+
+//A simple, intuitive, recursive approach
+//Time Complexity: O(n)
+//Space Complexity:: O(n)