[Feature Request]: leetcode problem 235-236 #934

dsa-solutions/lc-solutions/0200-0299/0235-lowest-common-Ancestor-of-binary-search-tree.md

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+---
+id: lowest-common-ancestor-of-a-binary-search-tree.
+title: Lowest Common Ancestor of a Binary Search Tree.
+sidebar_label: 235-Lowest-Common-Ancestor-of-a-Binary-Search-Tree
+tags:
+- Binary Search Tree
+- Tree
+- Depth-First Search
+- Binary Tree
+description: "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST."
+---
+
+## Problem
+
+Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
+
+According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
+
+### Examples
+
+**Example 1:**
+
+![image](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)
+
+**Input:** `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8`  
+**Output:** `6`  
+**Explanation:** The LCA of nodes 2 and 8 is 6.
+
+**Example 2:**
+
+![image](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)
+
+**Input:** `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4`  
+**Output:** `2`  
+**Explanation:** The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
+
+**Example 3:**
+
+**Input:** `root = [2,1], p = 2, q = 1`  
+**Output:** `2`
+
+### Constraints
+
+- The number of nodes in the tree is in the range `[2, 10^5]`.
+- `-10^9 <= Node.val <= 10^9`
+- All `Node.val` are unique.
+- `p != q`
+- `p` and `q` will exist in the BST.
+
+---
+
+## Approach
+
+To find the lowest common ancestor in a Binary Search Tree (BST), we can utilize the properties of the BST. The left subtree of a node contains only nodes with values less than the node's value, and the right subtree contains only nodes with values greater than the node's value.
+
+### Steps:
+
+1. **Start from the Root:** Begin the search from the root node of the BST.
+2. **Value Comparison:**
+   - If both `p` and `q` are smaller than the current node's value, move to the left child.
+   - If both `p` and `q` are greater than the current node's value, move to the right child.
+   - If `p` and `q` lie on either side of the current node, or one of them is the current node, then the current node is their LCA.
+
+### Solution
+
+#### Java
+
+```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 lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        while (root != null) {
+            // If both p and q are lesser than root, LCA must be in the left subtree
+            if (p.val < root.val && q.val < root.val) {
+                root = root.left;
+            }
+            // If both p and q are greater than root, LCA must be in the right subtree
+            else if (p.val > root.val && q.val > root.val) {
+                root = root.right;
+            }
+            // If p and q lie on either side of root, or one of them is the root, then root is the LCA
+            else {
+                return root;
+            }
+        }
+        return null;
+    }
+}
+```
+#### CPP
+```cpp
+/**
+ * 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* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        while (root != nullptr) {
+            // If both p and q are lesser than root, LCA must be in the left subtree
+            if (p->val < root->val && q->val < root->val) {
+                root = root->left;
+            }
+            // If both p and q are greater than root, LCA must be in the right subtree
+            else if (p->val > root->val && q->val > root->val) {
+                root = root->right;
+            }
+            // If p and q lie on either side of root, or one of them is the root, then root is the LCA
+            else {
+                return root;
+            }
+        }
+        return nullptr;
+    }
+};
+```
+### Python
+```python
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+class Solution:
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        while root:
+            # If both p and q are lesser than root, LCA must be in the left subtree
+            if p.val < root.val and q.val < root.val:
+                root = root.left
+            # If both p and q are greater than root, LCA must be in the right subtree
+            elif p.val > root.val and q.val > root.val:
+                root = root.right
+            # If p and q lie on either side of root, or one of them is the root, then root is the LCA
+            else:
+                return root
+        return None
+```    
+
+### Complexity Analysis
+#### Time Complexity: O(h)
+
+**Reason**: The algorithm may traverse the height h of the tree. In the worst case, this is O(log n) for a balanced BST and O(n) for a skewed BST.
+Space Complexity: O(1)
+
+**Reason**: The algorithm uses constant space.
+#### References
+**LeetCode Problem** : Lowest Common Ancestor of a Binary Search Tree
+**Solution Link**: LCA Solution on LeetCode
+
+**Wikipedia Definition**: Lowest Common Ancestor

dsa-solutions/lc-solutions/0200-0299/0236-Lowest-Common-Ancestor-of-a-Binary-Tree.md

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+---
+id: lowest-common-ancestor-of-a-binary-tree
+title: Lowest Common Ancestor of a Binary Tree
+sidebar_label: 0236-Lowest-Common-Ancestor-of-a-Binary-Tree
+tags:
+- Tree
+- Depth-First Search
+- Binary Tree
+description: "Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree."
+---
+
+## Problem
+
+Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
+
+According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
+
+### Examples
+
+**Example 1:**
+
+**Input:** `root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1`  
+**Output:** `3`  
+**Explanation:** The LCA of nodes 5 and 1 is 3.
+
+**Example 2:**
+
+**Input:** `root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4`  
+**Output:** `5`  
+**Explanation:** The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
+
+**Example 3:**
+
+**Input:** `root = [1,2], p = 1, q = 2`  
+**Output:** `1`
+
+### Constraints
+
+- The number of nodes in the tree is in the range `[2, 10^5]`.
+- `-10^9 <= Node.val <= 10^9`
+- All `Node.val` are unique.
+- `p != q`
+- `p` and `q` will exist in the tree.
+
+---
+
+## Approach
+
+To find the lowest common ancestor in a binary tree, we can use a recursive depth-first search (DFS) approach. The idea is to traverse the tree starting from the root. If we find either of the nodes `p` or `q`, we return that node. If both nodes are found in different subtrees of a node, then that node is their lowest common ancestor.
+
+### Steps:
+
+1. **Base Case:** If the current node is `null` or matches `p` or `q`, return the current node.
+2. **Recursive Search:**
+   - Recursively search the left subtree for `p` and `q`.
+   - Recursively search the right subtree for `p` and `q`.
+3. **Determine LCA:**
+   - If both left and right recursive calls return non-null, the current node is the LCA.
+   - If only one of the recursive calls returns non-null, that means both nodes are located in the same subtree, so return the non-null result.
+
+### Solution
+
+#### Java
+
+```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 lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        if (root == null || root == p || root == q) {
+            return root;
+        }
+        TreeNode left = lowestCommonAncestor(root.left, p, q);
+        TreeNode right = lowestCommonAncestor(root.right, p, q);
+        if (left != null && right != null) {
+            return root;
+        }
+        return left != null ? left : right;
+    }
+}
+```
+### C++
+```cpp
+/**
+ * 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* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        if (root == nullptr || root == p || root == q) {
+            return root;
+        }
+        TreeNode* left = lowestCommonAncestor(root->left, p, q);
+        TreeNode* right = lowestCommonAncestor(root->right, p, q);
+        if (left != nullptr && right != nullptr) {
+            return root;
+        }
+        return left != nullptr ? left : right;
+    }
+};
+```
+### Python
+```
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+class Solution:
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        if not root or root == p or root == q:
+            return root
+        left = self.lowestCommonAncestor(root.left, p, q)
+        right = self.lowestCommonAncestor(root.right, p, q)
+        if left and right:
+            return root
+        return left if left else right
+```
+### Complexity Analysis
+#### Time Complexity: O(n)
+
+**Reason:** The algorithm visits each node in the tree once, where n is the number of nodes in the tree.
+Space Complexity: O(h)
+
+**Reason:** The space complexity is determined by the height h of the tree due to the recursion stack. In the worst case, the height of the tree is O(n) for a skewed tree, but O(log n) for a balanced tree.
+
+### References
+**LeetCode Problem:** Lowest Common Ancestor of a Binary Tree
+
+**Solution Link:** LCA Solution on LeetCode
+**Wikipedia Definition:** Lowest Common Ancestor