Find distance between two nodes of a Binary Search Tree

Given a Binary Search Tree and two keys in it. Find the distance between two nodes with given two keys. It may be assumed that both keys exist in BST.


Input :  Root of above tree
         a = 3, b = 9
Output : 4
Distance between 3 and 9 in 
above BST is 4.

Input :  Root of above tree
         a = 9, b = 25
Output : 3
Distance between 9 and 25 in 
above BST is 3.

We have discussed distance between two nodes in binary tree. The time complexity of this solution is O(n)

In case of BST, we can find distance faster. We start from root and for every node, we do following.

  1. If both keys are greater than current node, we move to right child of current node.
  2. If both keys are smaller than current node, we move to left child of current node.
  3. If one keys is smaller and other key is greater, current node is Lowest Common Ancestor (LCA) of two nodes. We find distances of current node from two keys and return sum of the distances.
// CPP program to find distance between
// two nodes in BST
#include <bits/stdc++.h>
using namespace std;
struct Node {
    struct Node* left, *right;
    int key;
struct Node* newNode(int key)
    struct Node* ptr = new Node;
    ptr->key = key;
    ptr->left = ptr->right = NULL;
    return ptr;
// Standard BST insert function
struct Node* insert(struct Node* root, int key)
    if (!root)
        root = newNode(key);
    else if (root->key > key)
        root->left = insert(root->left, key);
    else if (root->key < key)
        root->right = insert(root->right, key);
    return root;
// This function returns distance of x from
// root. This function assumes that x exists
// in BST and BST is not NULL.
int distanceFromRoot(struct Node* root, int x)
    if (root->key == x)
        return 0;
    else if (root->key > x)
        return 1 + distanceFromRoot(root->left, x);
    return 1 + distanceFromRoot(root->right, x);
// Returns minimum distance beween a and b.
// This function assumes that a and b exist
// in BST.
int distanceBetween2(struct Node* root, int a, int b)
    if (!root)
        return 0;
    // Both keys lie in left
    if (root->key > a && root->key > b)
        return distanceBetween2(root->left, a, b);
    // Both keys lie in right
    if (root->key < a && root->key < b) // same path
        return distanceBetween2(root->right, a, b);
    // Lie in opposite directions (Root is
    // LCA of two nodes)
    if (root->key >= a && root->key <= b)
        return distanceFromRoot(root, a) +
               distanceFromRoot(root, b);
// This function make sure that a is smaller
// than b before making a call to findDistWrapper()
int findDistWrapper(Node *root, int a, int b)
   if (a > b)
     swap(a, b);
   return distanceBetween2(root, a, b);  
// Driver code
int main()
    struct Node* root = NULL;
    root = insert(root, 20);
    insert(root, 10);
    insert(root, 5);
    insert(root, 15);
    insert(root, 30);
    insert(root, 25);
    insert(root, 35);
    int a = 5, b = 55;
    cout << findDistWrapper(root, 5, 35);
    return 0;



Time Complexity : O(h) where h is height of Binary Search Tree.

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:
We have built the accelerating growth-oriented website for budding engineers and aspiring job holders of technology companies such as Google, Facebook, and Amazon
If you would like to study our free courses you can join us at #techcodebit #google #microsoft #facebook #interview portal #jobplacements


Leave a Reply

Your email address will not be published. Required fields are marked *

Skip to toolbar