You have been given a binary search tree of integers with ‘N’ nodes. You are also given 'KEY' which represents data of a node of this tree.
Your task is to return the predecessor and successor of the given node in the BST.
1. The predecessor of a node in BST is that node that will be visited just before the given node in the inorder traversal of the tree. If the given node is visited first in the inorder traversal, then its predecessor is NULL.
2. The successor of a node in BST is that node that will be visited immediately after the given node in the inorder traversal of the tree. If the given node is visited last in the inorder traversal, then its successor is NULL.
3. The node for which the predecessor and successor will not always be present. If not present, you can hypothetically assume it's position (Given that it is a BST) and accordingly find out the predecessor and successor.
4. A binary search tree (BST) is a binary tree data structure which has the following properties.
• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.
The first line contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example for further clarification.
The second line contains 'KEY' representing the data of the node for which the predecessor and successor are to be found.
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.
For example, the input for the tree depicted in the below image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level.
The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
The only line contains two single space-separated integers representing data values of the predecessor and the successor node, respectively. If any of the two doesn’t exist, print -1 in place of it.
You are not required to print anything explicitly. It has already been taken care of. Just implement the function.
15 10 20 8 12 16 25 -1 -1 -1 -1 -1 -1 -1 -1
10
8 12
The tree can be represented as follows:
The inorder traversal of this tree will be 8 10 12 15 16 20 25.
Since the node with data 8 is on the immediate left of the node with data 10 in the inorder traversal, the node with data 8 is the predecessor.
Since the node with data 12 is on the immediate right of the node with data 10 in the inorder traversal, the node with data 12 is the successor.
10 5 -1 -1 -1
5
-1 10
1 <= N <= 10^4
1 <= data <= 10^7
Time Limit: 1 sec
Store the nodes in the order of inorder traversal.
O(N), where ‘N’ is the number of nodes in the BST.
As we are traversing each node of the BST once, the time complexity will be linear.
O(N), where ‘N’ is the number of nodes in the BST.
In the worst case (skewed trees), we will have all the nodes of the BST in the recursion stack.
Also, the maximum possible size of the array used to store inorder traversal will be equal to N. Hence, the space complexity is linear.
All tags
Sort by
Interview problems
Python
def predecessorSuccessor(root, key):
pred=-1
succ=-1
def Predecessor(node):
if node is None:
return None
while node.right:
node=node.right
return node.data
def Successor(node):
if node is None:
return None
while node.left:
node=node.left
return node.data
curr=root
while curr:
if curr.data == key:
if curr.left:
pred=Predecessor(curr.left)
if curr.right:
succ=Successor(curr.right)
break
elif curr.data > key:
succ=curr.data
curr=curr.left
else:
pred=curr.data
curr=curr.right
return [pred,succ]
Interview problems
Predecessor And Successor In BST using binary search
void inorder(TreeNode *root,vector<int>&v)
{
if(root==NULL)
{
return;
}
inorder(root->left,v);
v.push_back(root->data);
inorder(root->right,v);
}
pair<int, int> predecessorSuccessor(TreeNode *root, int key)
{
// Write your code here.
vector<int>v;
inorder(root,v);
int s=0;
int e=v.size()-1;
int mid;
while(s<=e)
{
if(v[0]==key)
{
return make_pair(-1,v[1]);
}
if(v[0]>key)
{
return make_pair(-1,v[0]);
}
if(v[e]==key)
{
return make_pair(v[e-1],-1);
}
if(v[e]<key)
{
return make_pair(v[e],-1);
}
mid=s+(e-s)/2;
if(v[mid]==key)
{
return make_pair(v[mid-1],v[mid+1]);
}
if(v[mid]<key)
{
s=mid+1;
}
if(v[mid]<key && v[mid+1]>key)
{
return make_pair(v[mid],v[mid+1]);
}
if(v[mid]>key)
{
e=mid-1;
}
if(v[mid]>key && v[mid-1]<key)
{
return make_pair(v[mid-1],v[mid]);
}
}
}
Interview problems
Easy C++ solution to pass all test case
pair<int, int> predecessorSuccessor(TreeNode *root, int key)
{
// Write your code here.
if(root == NULL){
pair<int , int> p = make_pair(NULL , NULL);
return p;
}
int pred = -1;
int succ = -1;
TreeNode* temp = root;
while(temp != NULL){
if(temp -> data > key){
succ = temp -> data;
temp = temp -> left;
}
else if(temp -> data == key){
break;
}
else{
pred = temp -> data;
temp = temp -> right;
}
}
if(temp == NULL){
return make_pair(pred, succ);
}
TreeNode* leftNode = temp -> left;
while(leftNode != NULL){
pred = leftNode -> data;
leftNode = leftNode -> right;
}
TreeNode* rightNode = temp -> right;
while(rightNode != NULL){
succ = rightNode -> data;
rightNode = rightNode -> left;
}
return make_pair(pred, succ);
}
Interview problems
Wrong Test Case
6521171 5650934 9278816 899375 -1 8989449 -1 -1 -1 -1 -1 9615950 *9615950 is not present in BST
Interview problems
My Approach : ) { T.C. :- O(n) S.C. :- O(n) }
/*************************************************************
Following is the Binary Tree node structure
class TreeNode
{
public:
int data;
TreeNode *left, *right;
TreeNode() : data(0), left(NULL), right(NULL) {}
TreeNode(int x) : data(x), left(NULL), right(NULL) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : data(x), left(left), right(right) {}
};
*************************************************************/
TreeNode* insert_into_tree(TreeNode* &root, int key, bool &flag){
if(root == NULL && flag == 0){
TreeNode* temp = new TreeNode(key);
return temp;
}
if(root->data == key && flag == 0){
flag = 1;
return root;
}
if(root->data > key && flag == 0){
root->left = insert_into_tree(root->left, key, flag);
}
if(root->data < key && flag == 0){
root->right = insert_into_tree(root->right, key, flag);
}
return root; // forgot this before...!! (took too much time/ frustated me !! [ :( ]) (remember this from next time please...)
}
void inorder(TreeNode* root, vector<int> &v){
if(root == NULL){
return;
}
inorder(root->left, v);
v.push_back(root->data);
inorder(root->right, v);
}
int binary_search(int key, vector<int> &v){
int s = 0, e = v.size()-1;
int mid = s + (e-s)/2;
while(s <= e){
if(v[mid] == key){
return mid;
}
else if(v[mid] < key){
s = mid+1;
}
else{ // v[mid] > key
e = mid-1;
}
mid = s + (e-s)/2;
}
return -1;
}
pair<int, int> predecessorSuccessor(TreeNode *root, int key)
{
// Write your code here.
bool flag = 0;
insert_into_tree(root, key, flag);
vector<int> v; // will be sorted
inorder(root, v);
int index = binary_search(key, v);
pair<int, int> p;
if(v.size() == 1){
p.first = -1;
p.second = -1;
}
else if(index == 0 && v.size() > 1){
p.first = -1;
p.second = v[1];
}
else if(index == v.size()-1 && v.size() > 1){
p.first = v[v.size()-2];
p.second = -1;
}
else{
p.first = v[index-1];
p.second = v[index+1];
}
return p;
}
Interview problems
C++ ,T.C = O(n) , S.C = O(1);
int min_value(TreeNode *root){
TreeNode *temp = root;
while(temp->left!=NULL){
temp=temp->left;
}
return temp->data;
}
int max_value(TreeNode *root){
TreeNode *temp = root;
while(temp->right!=NULL){
temp=temp->right;
}
return temp->data;
}
pair<int, int> predecessorSuccessor(TreeNode *root, int key)
{
// Write your code here.
pair<int,int> answer;
answer.first = -1;
answer.second =-1;
TreeNode *temp = root;
while(temp != NULL && temp->data != key){
if(temp->data > key){
answer.second=temp->data;
temp=temp->left;
}
else{
answer.first=temp->data;
temp=temp->right;
}
}
if (temp == NULL) {
// Key not found in the tree
return answer;
}
if (temp->left != NULL) {
answer.first = max_value(temp->left);
}
if (temp->right != NULL) {
answer.second = min_value(temp->right);
}
return answer;
}
Interview problems
question jala do not found ka meaning ni aata
testcase 3
Interview problems
Easy to understand CPP solution.
This solution is easy to understand but it's not the best of the solutions. The problem statement in this question is not properly defined , there exists cases where the key itself is not present in the tree. Solution :
pair<int, int> predecessorSuccessor(TreeNode *root, int key)
{
// Write your code here.
if(root == NULL){
pair<int , int> p = make_pair(NULL , NULL);
return p;
}
int pred =-1;
int succ = -1;
TreeNode* temp = root;
while(temp!=NULL){
if(key>temp->data ){
pred = temp->data;
temp = temp->right;
}
else if(key<temp->data){
succ =temp->data;
temp = temp->left;
}
else if(key==temp->data){
break;
}
}
if(temp==NULL){
return make_pair(pred,succ);
}
TreeNode* leftTree = temp->left;
TreeNode* rightTree = temp->right;
while(leftTree!=NULL){
pred = leftTree->data;
leftTree = leftTree->right;
}
while(rightTree!=NULL){
succ = rightTree->data;
rightTree = rightTree->left;
}
return make_pair(pred,succ);
}
Interview problems
Solution Of Problem With TestCases:
Actually Problem is not with TestCases, Problem Statement Is Not Defined Properly,
Here You Will Not Get Key Everytime In BST and in the case where you do not find it, you need to return predecessor and successor of nearest value to that key,just like ceil and floor of a number
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
public class Solution {
public static List<Integer> ps(TreeNode root,int key,List<Integer>ans){
if(root==null) return ans;
if(root.data==key){
if(root.left!=null){
ans.set(0,root.left.data);
}
if(root.right!=null){
ans.set(1,root.right.data);
}
return ans;
}
if(root.data>key){
if(ans.get(1)>root.data || ans.get(1)==-1) ans.set(1,root.data);
return ps(root.left, key, ans);
}
if(root.data<key){
if(ans.get(0)<root.data || ans.get(0)==-1) ans.set(0,root.data);
return ps(root.right, key, ans);
}
return ans;
}
public static List<Integer> predecessorSuccessor(TreeNode root, int key) {
List<Integer>ans=new ArrayList<>();
ans.add(-1);
ans.add(-1);
return ps(root,key,ans);
}
}Interview problems
Most Optimized Java Solution T.C O(h)
import java.util.*;
public class Solution {
public static List<Integer> predecessorSuccessor(TreeNode root, int key) {
TreeNode successor = null;
TreeNode predcessor = null;
TreeNode node = root;
while (node != null) {
// System.out.println(node.data);
if(node.data > key){
successor = node;
node = node.left;
}else{
node = node.right;
}
}
node = root;
// System.out.println("-------");
while (node != null) {
// System.out.println(node.data);
if(node.data >=key){
node = node.left;
}else{
predcessor = node;
node = node.right;
// break;
}
}
List<Integer> ans = new ArrayList<>();
ans.add((predcessor == null ? -1 : predcessor.data));
ans.add(successor == null ? -1 : successor.data);
return ans;
}
}
