Algorithms‎ > ‎Trees‎ > ‎BST‎ > ‎

LCA

Lowest Common Ancestor in a Binary Search Tree.

August 9, 2009

Given the values of two nodes in a *binary search tree*, write a c program to find the lowest common ancestor. You may assume that both values already exist in the tree.

The function prototype is as follows:

 int FindLowestCommonAncestor(node* root, int value1, int value)

BST_LCA

  I/P : 4 and 14
  O/P : 8
  (Here the common ancestors of 4 and 14, are {8,20}.
  Of {8,20}, the lowest one is 8).

Here is the solution

Algorithm:
The main idea of the solution is — While traversing Binary Search Tree from top to bottom, the first node n we encounter with value between n1 and n2, i.e., n1 < n < n2 is the Lowest or Least Common Ancestor(LCA) of n1 and n2 (where n1 < n2). So just traverse the BST in pre-order, if you find a node with value in between n1 and n2 then n is the LCA, if it's value is greater than both n1 and n2 then our LCA lies on left side of the node, if it's value is smaller than both n1 and n2 then LCA lies on right side.

Implementation:

#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node
{
    int data;
    struct node* left;
    struct node* right;
};
 
struct node* newNode(int );
 
/* Function to find least comman ancestor of n1 and n2 */
int leastCommanAncestor(struct node* root, int n1, int n2)
{
  /* If we have reached a leaf node then LCA doesn't exist
     If root->data is equal to any of the inputs then input is
     not valid. For example 20, 22 in the given figure */ 
  if(root == NULL || root->data == n1 || root->data == n2)
    return -1;
   
  /* If any of the input nodes is child of the current node
     we have reached the LCA. For example, in the above figure
     if we want to calculate LCA of 12 and 14, recursion should
     terminate when we reach 8*/
  if((root->right != NULL) &&
    (root->right->data == n1 || root->right->data == n2))
    return root->data;
  if((root->left != NULL) &&
    (root->left->data == n1 || root->left->data == n2))
    return root->data;   
     
  if(root->data > n1 && root->data < n2)
    return root->data;
  if(root->data > n1 && root->data > n2)
    return leastCommanAncestor(root->left, n1, n2);
  if(root->data < n1 && root->data < n2)
    return leastCommanAncestor(root->right, n1, n2);
}   
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
  struct node* node = (struct node*)
                       malloc(sizeof(struct node));
  node->data  = data;
  node->left  = NULL;
  node->right = NULL;
 
  return(node);
}
 
/* Driver program to test mirror() */
int main()
{
  struct node *root  = newNode(2);
  root->left         = newNode(1);
  root->right        = newNode(4);
  root->right->left  = newNode(3);
  root->right->right = newNode(5);
 
/* Constructed binary search tree is
            2
           / \
         1   4
             / \
           3   5
*/
  printf("\n The Least Common Ancestor is \n");
  printf("%d", leastCommanAncestor(root, 3, 5));
 
  getchar();
  return 0;
}

Note that above function assumes that n1 is smaller than n2.

Time complexity: Time complexity is O(Logn) for a balanced BST and O(n) for a skewed BST.

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