Lowest Common Ancestor in a Binary Tree

Author: spaceman-dev

Some_More_Trees/LowestCommonAncestorInABinaryTree.cpp

@@ -0,0 +1,153 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// Tree Node
+struct Node
+{
+ int data;
+ Node* left;
+ Node* right;
+};
+
+ // } Driver Code Ends
+/* A binary tree node
+
+struct Node
+{
+ int data;
+ struct Node* left;
+ struct Node* right;
+ 
+ Node(int x){
+ data = x;
+ left = right = NULL;
+ }
+};
+ */
+
+class Solution
+{
+ public:
+ //Function to return the lowest common ancestor in a Binary Tree.
+ Node* lca(Node* root ,int n1 ,int n2 )
+ {
+ //Your code here 
+ if(!root)
+ return NULL;
+ if(root->data == n1 || root->data == n2)
+ return root;
+ Node* left = lca(root->left, n1, n2);
+ Node* right = lca(root->right, n1, n2);
+ if(left && right)
+ return root;
+ return left ? left : right;
+ }
+};
+
+// { Driver Code Starts.
+
+Node* newNode(int val)
+{
+ Node* temp = new Node;
+ temp->data = val;
+ temp->left = NULL;
+ temp->right = NULL;
+
+ return temp;
+}
+
+
+// Function to Build Tree
+Node* buildTree(string str)
+{
+ // Corner Case
+ if(str.length() == 0 || str[0] == 'N')
+ return NULL;
+
+ // Creating vector of strings from input
+ // string after spliting by space
+ vector<string> ip;
+
+ istringstream iss(str);
+ for(string str; iss >> str; )
+ ip.push_back(str);
+
+ // for(string i:ip)
+ // cout<<i<<" ";
+ // cout<<endl;
+ // Create the root of the tree
+ Node* root = newNode(stoi(ip[0]));
+
+ // Push the root to the queue
+ queue<Node*> queue;
+ queue.push(root);
+
+ // Starting from the second element
+ int i = 1;
+ while(!queue.empty() && i < ip.size()) {
+
+ // Get and remove the front of the queue
+ Node* currNode = queue.front();
+ queue.pop();
+
+ // Get the current node's value from the string
+ string currVal = ip[i];
+
+ // If the left child is not null
+ if(currVal != "N") {
+
+ // Create the left child for the current node
+ currNode->left = newNode(stoi(currVal));
+
+ // Push it to the queue
+ queue.push(currNode->left);
+ }
+
+ // For the right child
+ i++;
+ if(i >= ip.size())
+ break;
+ currVal = ip[i];
+
+ // If the right child is not null
+ if(currVal != "N") {
+
+ // Create the right child for the current node
+ currNode->right = newNode(stoi(currVal));
+
+ // Push it to the queue
+ queue.push(currNode->right);
+ }
+ i++;
+ }
+
+ return root;
+}
+
+// Function for Inorder Traversal
+void printInorder(Node* root)
+{
+ if(!root)
+ return;
+
+ printInorder(root->left);
+ cout<<root->data<<" ";
+ printInorder(root->right);
+}
+
+int main() {
+ int t;
+ scanf("%d",&t);
+ while(t--)
+ {
+ int a,b;
+ scanf("%d %d ",&a,&b);
+ string s;
+ getline(cin,s);
+ Node* root = buildTree(s);
+ Solution ob;
+ cout<<ob.lca(root,a,b)->data<<endl;
+ }
+ return 0;
+}
+ // } Driver Code Ends

Some_More_Trees/README.md

@@ -50,4 +50,14 @@
   / \
   3 2
  Output: 3 1 2
- Explanation: First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 3 and right child of 1 is 2.
+ Explanation: First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 3 and right child of 1 is 2.
+
+## 4. Lowest Common Ancestor in a Binary Tree:
+ Given a Binary Tree with all unique values and two nodes value n1 and n2. The task is to find the lowest common ancestor of the given two nodes. We may assume that either both n1 and n2 are present in the tree or none of them is present. 
+ Example 1:
+ Input: n1 = 2 , n2 = 3
+  1
+  / \
+ 2 3
+ Output: 1
+ Explanation: LCA of 2 and 3 is 1.