Added docs for lowest common ancestor problem in DSA

docs/DSA-Problem-Solution/lowest_common_ancestor.md

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+---
+id: Lowest-Common-Ancestor
+title: Lowest Common Ancestor
+sidebar_label: Lowest Common Ancestor
+sidebar_position: 12
+description: A detailed 
+tags: [lowest common ancestor, lca, dsa, prolem solving]
+---
+
+# Lowest Common Ancestor (LCA) in a Binary Tree
+
+The **Lowest Common Ancestor (LCA)** of two nodes `p` and `q` in a binary tree is defined as the lowest node in the tree that has both `p` and `q` as descendants (where a node can be a descendant of itself).
+
+---
+
+## Problem Statement
+
+Given a binary tree and two nodes `p` and `q`, find their lowest common ancestor.
+
+### Node Class Representation
+
+**C++ Class Definition**:
+
+```cpp
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+};
+```
+
+**Python Class Definition**:
+
+```python
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+```
+
+---
+
+## Solution Approach: Recursive Strategy
+
+We use a **recursive approach** to solve the LCA problem efficiently. The idea is to traverse the tree and return the node if it matches either `p` or `q`. Otherwise, we check the left and right subtrees for `p` and `q` and determine the LCA based on the results.
+
+### C++ Implementation
+
+```cpp
+#include <iostream>
+using namespace std;
+
+// Definition for a binary tree node
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+};
+
+class Solution {
+public:
+    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        if (!root || root == p || root == q) return root; // Base case
+
+        TreeNode* left = lowestCommonAncestor(root->left, p, q);
+        TreeNode* right = lowestCommonAncestor(root->right, p, q);
+
+        if (left && right) return root; // If both left and right are not null, root is the LCA
+        return left ? left : right;     // Otherwise, return the non-null child
+    }
+};
+```
+
+### Python Implementation
+
+```python
+class Solution:
+    def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
+        if not root or root == p or root == q:  # Base case
+            return root
+
+        left = self.lowestCommonAncestor(root.left, p, q)
+        right = self.lowestCommonAncestor(root.right, p, q)
+
+        if left and right:
+            return root  # If both left and right are not null, root is the LCA
+        return left if left else right  # Otherwise, return the non-null child
+```
+
+---
+
+### Time Complexity: `O(N)`
+
+- `N` is the number of nodes in the binary tree.
+- We traverse each node of the binary tree only once.
+
+### Space Complexity: `O(H)`
+
+- `H` is the height of the tree, which represents the space used by the recursion stack.
+- In the worst case (a skewed tree), `H` can be `O(N)`. In a balanced tree, `H` is `O(log N)`.
+
+---
+
+## Key Concepts to Understand
+
+1. **Binary Tree Traversal**: Understanding how to traverse the binary tree recursively.
+2. **Recursive Approach**: Using recursion to check for `p` and `q` in both left and right subtrees.
+3. **Base Cases**: Properly handling cases where the current node is `p` or `q`, or if the current node is `nullptr`.
+
+---
+
+## Solution Approach: Iterative Strategy (Optional)
+
+For cases where recursion is not ideal, an **iterative approach** using parent pointers and a set can be used. Here’s a brief outline for this approach:
+
+1. Use a **hash map** to store parent pointers of all nodes in the tree.
+2. Use this information to track ancestors of `p` and then find the first common ancestor of `q`.
+
+### Python Implementation (Iterative)
+
+```python
+def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
+    parent = {root: None}
+    stack = [root]
+
+    # Traverse the tree until both nodes are found
+    while p not in parent or q not in parent:
+        node = stack.pop()
+        if node.left:
+            parent[node.left] = node
+            stack.append(node.left)
+        if node.right:
+            parent[node.right] = node
+            stack.append(node.right)
+
+    # Store ancestors of p in a set
+    ancestors = set()
+    while p:
+        ancestors.add(p)
+        p = parent[p]
+
+    # Find the first common ancestor of q
+    while q not in ancestors:
+        q = parent[q]
+
+    return q
+```
+
+### Time Complexity: `O(N)`
+
+- Traversing the tree to fill the parent pointers takes `O(N)` time.
+- Tracing back the ancestors of `p` and `q` also takes `O(N)` time.
+
+### Space Complexity: `O(N)`
+
+- The hash map and ancestor set both use `O(N)` space.
+
+---
+
+## Summary
+
+- The **LCA Problem** is a fundamental question in tree data structures, often appearing in coding interviews.
+- The **recursive solution** is simple and efficient for balanced binary trees.
+- The **iterative solution** is useful when recursion depth might become a concern.
+
+Mastering this problem will strengthen your understanding of binary tree traversal and recursion, making it easier to solve more complex tree-based problems. Happy coding!