Lowest Common Ancestor in a Binary Tree

Problem Statement - link #

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 are present.

LCA: It is the first common ancestor of both the nodes n1 and n2 from bottom of tree.

Your Task: You don’t have to read, input, or print anything. Your task is to complete the function lca() that takes nodes, n1, and n2 as parameters and returns the LCA node as output.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(h) , where h = height of tree

Examples #

Example 1:

Input:
n1 = 2 , n2 = 3  
       1 
      / \ 
     2   3
Output: 1
Explanation:
LCA of 2 and 3 is 1.

Example 2:

Input:
n1 = 3 , n2 = 4
           5    
          /    
         2  
        / \  
       3   4
Output: 2
Explanation:
LCA of 3 and 4 is 2. 

Constraints #

• 1 <= Number of nodes <= 10^5
• 1 <= Data of a node <= 10^5

Solutions #

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 root;
       if(root->data==n1 || root->data==n2) return root;
       Node* pre = lca(root->left,n1,n2);
       Node* nex = lca(root->right,n1,n2);
       if(pre && nex) return root;
       return pre ? pre : nex;
    }
};