Inorder predecessor and successor for a given key in BST

• Difficulty Level : Medium
• Last Updated : 29 Jul, 2021

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:

C++

struct Node
{
int key;
struct Node *left, *right ;
};

Java

static class Node
{
int key;
Node left, right ;
};

// This code is contributed by gauravrajput1

Python3

class Node:

def __init__(self, key):

self.key = key
self.left = None
self.right = None

# This code is contributed by harshitkap00r

C#

public class Node
{
public int key;
public Node left, right ;
};

// This code is contributed by gauravrajput1

Javascript

<script>

class Node {
constructor() {
this.key = 0;
this.left = null;
this.right = null;
}
}

</script>

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

Following is the implementation of the above algorithm:

C++

// 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;
}

Java

// Java program to find predecessor
// and successor in a BST
class GFG{

// BST Node
static class Node
{
int key;
Node left, right;

public Node()
{}

public Node(int key)
{
this.key = key;
this.left = this.right = null;
}
};

static Node pre = new Node(), suc = new Node();

// This function finds predecessor and
// successor of key in BST. It sets pre
// and suc as predecessor and successor
// respectively
static void findPreSuc(Node root, 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 != null)
tmp = tmp.right;

pre = tmp;
}

// The minimum value in
// right subtree is successor
if (root.right != null)
{
Node tmp = root.right;

while (tmp.left != null)
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, key);
}

// Go to right subtree
else
{
pre = root;
findPreSuc(root.right, key);
}
}

// A utility function to insert a
// new node with given key in BST
static Node insert(Node node, int key)
{
if (node == null)
return new Node(key);
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);

return node;
}

// Driver code
public static void main(String[] args)
{

// Key to be searched in BST
int key = 65;

/*
* Let us create following BST
*          50
*         /  \
*        30   70
*       /  \ /  \
*      20 40 60  80
*/

Node root = new Node();
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);

findPreSuc(root, key);
if (pre != null)
System.out.println("Predecessor is " + pre.key);
else
System.out.println("No Predecessor");

if (suc != null)
System.out.println("Successor is " + suc.key);
else
System.out.println("No Successor");
}
}

// This code is contributed by sanjeev2552

Python

# Python program to find predecessor and successor in a BST

# A BST node
class Node:

# Constructor to create a new node
def __init__(self, key):
self.key  = key
self.left = None
self.right = None

# This function finds predecessor and successor of key in BST
# It sets pre and suc as predecessor and successor respectively
def findPreSuc(root, key):

# Base Case
if root is None:
return

# If key is present at root
if root.key == key:

# the maximum value in left subtree is predecessor
if root.left is not None:
tmp = root.left
while(tmp.right):
tmp = tmp.right
findPreSuc.pre = tmp

# the minimum value in right subtree is successor
if root.right is not None:
tmp = root.right
while(temp.left):
tmp = tmp.left
findPreSuc.suc = tmp

return

# If key is smaller than root's key, go to left subtree
if root.key > key :
findPreSuc.suc = root
findPreSuc(root.left, key)

else: # go to right subtree
findPreSuc.pre = root
findPreSuc(root.right, key)

# A utility function to insert a new node in with given key in BST
def insert(node , key):
if node is None:
return Node(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
key = 65 #Key to be searched in BST

""" Let us create following BST
50
/     \
30      70
/  \    /  \
20   40  60   80
"""
root = None
root = insert(root, 50)
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);

# Static variables of the function findPreSuc
findPreSuc.pre = None
findPreSuc.suc = None

findPreSuc(root, key)

if findPreSuc.pre is not None:
print "Predecessor is", findPreSuc.pre.key

else:
print "No Predecessor"

if findPreSuc.suc is not None:
print "Successor is", findPreSuc.suc.key
else:
print "No Successor"

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program to find predecessor
// and successor in a BST
using System;
public class GFG
{

// BST Node
public

class Node
{
public
int key;
public
Node left, right;
public Node()
{}

public Node(int key)
{
this.key = key;
this.left = this.right = null;
}
};

static Node pre = new Node(), suc = new Node();

// This function finds predecessor and
// successor of key in BST. It sets pre
// and suc as predecessor and successor
// respectively
static void findPreSuc(Node root, 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 != null)
tmp = tmp.right;

pre = tmp;
}

// The minimum value in
// right subtree is successor
if (root.right != null)
{
Node tmp = root.right;

while (tmp.left != null)
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, key);
}

// Go to right subtree
else
{
pre = root;
findPreSuc(root.right, key);
}
}

// A utility function to insert a
// new node with given key in BST
static Node insert(Node node, int key)
{
if (node == null)
return new Node(key);
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);

return node;
}

// Driver code
public static void Main(String[] args)
{

// Key to be searched in BST
int key = 65;

/*
* Let us create following BST
*          50
*         /  \
*        30   70
*       /  \ /  \
*      20 40 60  80
*/

Node root = new Node();
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);

findPreSuc(root, key);
if (pre != null)
Console.WriteLine("Predecessor is " + pre.key);
else
Console.WriteLine("No Predecessor");

if (suc != null)
Console.WriteLine("Successor is " + suc.key);
else
Console.WriteLine("No Successor");
}
}

// This code is contributed by aashish1995

Javascript

<script>

// JavaScript program to find predecessor
// and successor in a BST// BST Node
class Node
{
constructor(key)
{
this.key = key;
this.left = this.right = null;
}
}

var pre = new Node(), suc = new Node();

// This function finds predecessor and
// successor of key in BST. It sets pre
// and suc as predecessor and successor
// respectively
function findPreSuc(root , 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)
{
var tmp = root.left;
while (tmp.right != null)
tmp = tmp.right;

pre = tmp;
}

// The minimum value in
// right subtree is successor
if (root.right != null)
{
var tmp = root.right;

while (tmp.left != null)
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, key);
}

// Go to right subtree
else
{
pre = root;
findPreSuc(root.right, key);
}
}

// A utility function to insert a
// new node with given key in BST
function insert(node , key)
{
if (node == null)
return new Node(key);
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);

return node;
}

// Driver code

// Key to be searched in BST
var key = 65;

/*
* Let us create following BST
*          50
*         /  \
*        30   70
*       /  \ /  \
*      20 40 60  80
*/

var root = new Node();
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);

findPreSuc(root, key);
if (pre != null)
document.write("Predecessor is " + pre.key);
else
document.write("No Predecessor");

if (suc != null)
document.write("<br/>Successor is " + suc.key);
else
document.write("<br/>No Successor");

// This code contributed by gauravrajput1

</script>

Output:

Predecessor is 60
Successor is 70

Another Approach :
We can also find the inorder successor and inorder predecessor using inorder traversal . Check if the current node is smaller than the given key for predecessor and for successor, check if it is greater than the given key. If it is greater than the given key then, check if it is smaller than the already stored value in successor then, update it. At last, get the predecessor and successor stored in q(successor) and p(predecessor).

C++

// CPP code for inorder successor
// and predecessor of tree
#include<iostream>
#include<stdlib.h>

using namespace std;

struct Node
{
int data;
Node* left,*right;
};

// Function to return data
Node* getnode(int info)
{
Node* p = (Node*)malloc(sizeof(Node));
p->data = info;
p->right = NULL;
p->left = NULL;
return p;
}

/*
since inorder traversal results in
ascending order visit to node , we
can store the values of the largest
no which is smaller than a (predecessor)
and smallest no which is large than
a (successor) using inorder traversal
*/
void find_p_s(Node* root,int a,
Node** p, Node** q)
{
// If root is null return
if(!root)
return ;

// traverse the left subtree
find_p_s(root->left, a, p, q);

// root data is greater than a
if(root&&root->data > a)
{

// q stores the node whose data is greater
// than a and is smaller than the previously
// stored data in *q which is successor
if((!*q) || (*q) && (*q)->data > root->data)
*q = root;
}

// if the root data is smaller than
// store it in p which is predecessor
else if(root && root->data < a)
{
*p = root;
}

// traverse the right subtree
find_p_s(root->right, a, p, q);
}

// Driver code
int main()
{
Node* root1 = getnode(50);
root1->left = getnode(20);
root1->right = getnode(60);
root1->left->left = getnode(10);
root1->left->right = getnode(30);
root1->right->left = getnode(55);
root1->right->right = getnode(70);
Node* p = NULL, *q = NULL;

find_p_s(root1, 55, &p, &q);

if(p)
cout << p->data;
if(q)
cout << " " << q->data;
return 0;
}

Python3

""" Python3 code for inorder successor
and predecessor of tree """

# A Binary Tree Node
# Utility function to create a new tree node
class getnode:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None

"""
since inorder traversal results in
ascendingorder visit to node , we
can store the values of the largest
o which is smaller than a (predecessor)
and smallest no which is large than
a (successor) using inorder traversal
"""
def find_p_s(root, a, p, q):

# If root is None return
if(not root):
return

# traverse the left subtree
find_p_s(root.left, a, p, q)

# root data is greater than a
if(root and root.data > a):

# q stores the node whose data is greater
# than a and is smaller than the previously
# stored data in *q which is successor
if((not q) or q and
q.data > root.data):
q = root

# if the root data is smaller than
# store it in p which is predecessor
elif(root and root.data < a):
p= root

# traverse the right subtree
find_p_s(root.right, a, p, q)

# Driver Code
if __name__ == '__main__':

root1 = getnode(50)
root1.left = getnode(20)
root1.right = getnode(60)
root1.left.left = getnode(10)
root1.left.right = getnode(30)
root1.right.left = getnode(55)
root1.right.right = getnode(70)
p = [None]
q = [None]

find_p_s(root1, 55, p, q)

if(p) :
print(p.data, end = "")
if(q) :
print("", q.data)

# This code is contributed by
# SHUBHAMSINGH10

Javascript

<script>

class Node
{
constructor(data)
{
this.data = data;
this.left = this.right = null;
}
}

function find_p_s(root, a, p, q)
{
// If root is None return
if(root == null)
return

// traverse the left subtree
find_p_s(root.left, a, p, q)

// root data is greater than a
if(root && root.data > a)
{
// q stores the node whose data is greater
// than a and is smaller than the previously
// stored data in *q which is successor
if((q == null) || q != null && q.data > root.data)

q = root

}

// if the root data is smaller than
// store it in p which is predecessor
else if(root && root.data < a)
{    p = root

}

// traverse the right subtree
find_p_s(root.right, a, p, q)
}

// Driver Code
let root1 = new Node(50)
root1.left = new Node(20)
root1.right = new Node(60)
root1.left.left = new Node(10)
root1.left.right = new Node(30)
root1.right.left = new Node(55)
root1.right.right = new Node(70)
p = [null]
q = [null]

find_p_s(root1, 55, p, q)

if(p != null)
document.write(p.data, end = " ")
if(q != null)
document.write(" ", q.data)

// This code is contributed by patel2127
</script>

Output :

50 60

Thanks Shweta for suggesting this method.

?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk