I recently encountered with a question in an interview at e-commerce company. The interviewer asked the following question:
There is BST given with root node with key part as integer only. The structure of each node is as follows:
struct Node
{
int key;
struct Node *left, *right ;
};
You need to find the inorder successor and predecessor of a given key. In case the given key is not found in BST, then return the two values within which this key will lie.
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Following is the algorithm to reach the desired result. Its a recursive method:
Input: root node, key
output: predecessor node, successor node
1. If root is NULL
then return
2. if key is found then
a. If its left subtree is not null
Then predecessor will be the right most
child of left subtree or left child itself.
b. If its right subtree is not null
The successor will be the left most child
of right subtree or right child itself.
return
3. If key is smaller then root node
set the successor as root
search recursively into left subtree
else
set the predecessor as root
search recursively into right subtree
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Following is C++ implementation of the above algorithm:
C++ Programming
// C++ program to find predecessor and successor in a BST
#include <iostream>
using namespace std;
// BST Node
struct Node
{
int key;
struct Node *left, *right;
};
// This function finds predecessor and successor of key in BST.
// It sets pre and suc as predecessor and successor respectively
void findPreSuc(Node* root, Node*& pre, Node*& suc, int key)
{
// Base case
if (root == NULL) return ;
// If key is present at root
if (root->key == key)
{
// the maximum value in left subtree is predecessor
if (root->left != NULL)
{
Node* tmp = root->left;
while (tmp->right)
tmp = tmp->right;
pre = tmp ;
}
// the minimum value in right subtree is successor
if (root->right != NULL)
{
Node* tmp = root->right ;
while (tmp->left)
tmp = tmp->left ;
suc = tmp ;
}
return ;
}
// If key is smaller than root's key, go to left subtree
if (root->key > key)
{
suc = root ;
findPreSuc(root->left, pre, suc, key) ;
}
else // go to right subtree
{
pre = root ;
findPreSuc(root->right, pre, suc, key) ;
}
}
// A utility function to create a new BST node
Node *newNode(int item)
{
Node *temp = new Node;
temp->key = item;
temp->left = temp->right = NULL;
return temp;
}
/* A utility function to insert a new node with given key in BST */
Node* insert(Node* node, int key)
{
if (node == NULL) return newNode(key);
if (key < node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
return node;
}
// Driver program to test above function
int main()
{
int key = 65; //Key to be searched in BST
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
Node *root = NULL;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
Node* pre = NULL, *suc = NULL;
findPreSuc(root, pre, suc, key);
if (pre != NULL)
cout << "Predecessor is " << pre->key << endl;
else
cout << "No Predecessor";
if (suc != NULL)
cout << "Successor is " << suc->key;
else
cout << "No Successor";
return 0;
}
Output:
Predecessor is 60
Successor is 70