Inorder Successor in Binary Search Tree

In Binary Tree, Inorder successor of a node is the next node in Inorder traversal of the Binary Tree. Inorder Successor is NULL for the last node in Inoorder traversal.
In Binary Search Tree, Inorder Successor of an input node can also be defined as the node with the smallest key greater than the key of input node. So, it is sometimes important to find next node in sorted order.

In the above diagram, inorder successor of 8 is 10, inorder successor of 10 is 12 and inorder successor of 14 is 20.

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

Method 1 (Uses Parent Pointer)
In this method, we assume that every node has parent pointer.

The Algorithm is divided into two cases on the basis of right subtree of the input node being empty or not.

Input: node, root // node is the node whose Inorder successor is needed.
output: succ // succ is Inorder successor of node.

1) If right subtree of node is not NULL, then succ lies in right subtree. Do following.
Go to right subtree and return the node with minimum key value in right subtree.
2) If right sbtree of node is NULL, then succ is one of the ancestors. Do following.
Travel up using the parent pointer until you see a node which is left child of it’s parent. The parent of such a node is the succ.

Implementation
Note that the function to find InOrder Successor is highlighted (with gray background) in below code.

C

#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* parent; 
}; 
  
struct node * minValue(struct node* node);  
  
struct node * inOrderSuccessor(struct node *root, struct node *n) 
{ 
  // step 1 of the above algorithm  
  if( n->right != NULL ) 
    return minValue(n->right); 
  
  // step 2 of the above algorithm 
  struct node *p = n->parent; 
  while(p != NULL && n == p->right) 
  { 
     n = p; 
     p = p->parent; 
  } 
  return p; 
} 
  
/* Given a non-empty binary search tree, return the minimum data   
    value found in that tree. Note that the entire tree does not need 
    to be searched. */
struct node * minValue(struct node* node) { 
  struct node* current = node; 
   
  /* loop down to find the leftmost leaf */
  while (current->left != NULL) { 
    current = current->left; 
  } 
  return current; 
} 
  
/* 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; 
  node->parent = NULL; 
    
  return(node); 
} 
  
/* Give a binary search tree and a number, inserts a new node with     
    the given number in the correct place in the tree. Returns the new 
    root pointer which the caller should then use (the standard trick to  
    avoid using reference parameters). */
struct node* insert(struct node* node, int data) 
{ 
  /* 1. If the tree is empty, return a new, 
      single node */
  if (node == NULL) 
    return(newNode(data)); 
  else
  { 
    struct node *temp;   
  
    /* 2. Otherwise, recur down the tree */
    if (data <= node->data) 
    {     
         temp = insert(node->left, data); 
         node->left  = temp; 
         temp->parent= node; 
    } 
    else
    { 
        temp = insert(node->right, data); 
        node->right = temp; 
        temp->parent = node; 
    }     
   
    /* return the (unchanged) node pointer */
    return node; 
  } 
}  
   
/* Driver program to test above functions*/
int main() 
{ 
  struct node* root = NULL, *temp, *succ, *min; 
  
  //creating the tree given in the above diagram 
  root = insert(root, 20); 
  root = insert(root, 8); 
  root = insert(root, 22); 
  root = insert(root, 4); 
  root = insert(root, 12); 
  root = insert(root, 10);   
  root = insert(root, 14);     
  temp = root->left->right->right; 
  
  succ =  inOrderSuccessor(root, temp); 
  if(succ !=  NULL) 
    printf("\n Inorder Successor of %d is %d ", temp->data, succ->data);     
  else
    printf("\n Inorder Successor doesn't exit"); 
  
  getchar(); 
  return 0; 
} 

Java

// Java program to find minimum value node in Binary Search Tree 
  
// A binary tree node 
class Node { 
  
    int data; 
    Node left, right, parent; 
  
    Node(int d) { 
        data = d; 
        left = right = parent = null; 
    } 
} 
  
class BinaryTree { 
  
    static Node head; 
  
    /* Given a binary search tree and a number,  
     inserts a new node with the given number in  
     the correct place in the tree. Returns the new  
     root pointer which the caller should then use  
     (the standard trick to avoid using reference  
     parameters). */
    Node insert(Node node, int data) { 
  
        /* 1. If the tree is empty, return a new,      
         single node */
        if (node == null) { 
            return (new Node(data)); 
        } else { 
  
            Node temp = null; 
              
            /* 2. Otherwise, recur down the tree */
            if (data <= node.data) { 
                temp = insert(node.left, data); 
                node.left = temp; 
                temp.parent = node; 
  
            } else { 
                temp = insert(node.right, data); 
                node.right = temp; 
                temp.parent = node; 
            } 
  
            /* return the (unchanged) node pointer */
            return node; 
        } 
    } 
  
    Node inOrderSuccessor(Node root, Node n) { 
  
        // step 1 of the above algorithm  
        if (n.right != null) { 
            return minValue(n.right); 
        } 
  
        // step 2 of the above algorithm 
        Node p = n.parent; 
        while (p != null && n == p.right) { 
            n = p; 
            p = p.parent; 
        } 
        return p; 
    } 
  
    /* Given a non-empty binary search tree, return the minimum data   
     value found in that tree. Note that the entire tree does not need 
     to be searched. */
    Node minValue(Node node) { 
        Node current = node; 
  
        /* loop down to find the leftmost leaf */
        while (current.left != null) { 
            current = current.left; 
        } 
        return current; 
    } 
  
    // Driver program to test above functions 
    public static void main(String[] args) { 
        BinaryTree tree = new BinaryTree(); 
        Node root = null, temp = null, suc = null, min = null; 
        root = tree.insert(root, 20); 
        root = tree.insert(root, 8); 
        root = tree.insert(root, 22); 
        root = tree.insert(root, 4); 
        root = tree.insert(root, 12); 
        root = tree.insert(root, 10); 
        root = tree.insert(root, 14); 
        temp = root.left.right.right; 
        suc = tree.inOrderSuccessor(root, temp); 
        if (suc != null) { 
            System.out.println("Inorder successor of " + temp.data +  
                                                      " is " + suc.data); 
        } else { 
            System.out.println("Inorder successor does not exist"); 
        } 
    } 
} 
  
// This code has been contributed by Mayank Jaiswal 

Python

# Python program to find the inroder successor in a BST 
  
# A binary tree node  
class Node: 
  
    # Constructor to create a new node 
    def __init__(self, key): 
        self.data = key  
        self.left = None
        self.right = None
  
def inOrderSuccessor(root, n): 
      
    # Step 1 of the above algorithm 
    if n.right is not None: 
        return minValue(n.right) 
  
    # Step 2 of the above algorithm 
    p = n.parent 
    while( p is not None): 
        if n != p.right : 
            break 
        n = p  
        p = p.parent 
    return p 
  
# Given a non-empty binary search tree, return the  
# minimum data value found in that tree. Note that the 
# entire tree doesn't need to be searched 
def minValue(node): 
    current = node 
  
    # loop down to find the leftmost leaf 
    while(current is not None): 
        if current.left is None: 
            break
        current = current.left 
  
    return current 
  
  
# Given a binary search tree and a number, inserts a 
# new node with the given number in the correct place 
# in the tree. Returns the new root pointer which the 
# caller should then use( the standard trick to avoid  
# using reference parameters) 
def insert( node, data): 
  
    # 1) If tree is empty then return a new singly node 
    if node is None: 
        return Node(data) 
    else: 
         
        # 2) Otherwise, recur down the tree 
        if data <= node.data: 
            temp = insert(node.left, data) 
            node.left = temp  
            temp.parent = node 
        else: 
            temp = insert(node.right, data) 
            node.right = temp  
            temp.parent = node 
          
        # return  the unchanged node pointer 
        return node 
  
  
# Driver progam to test above function 
  
root  = None
  
# Creating the tree given in the above diagram  
root = insert(root, 20) 
root = insert(root, 8); 
root = insert(root, 22); 
root = insert(root, 4); 
root = insert(root, 12); 
root = insert(root, 10);   
root = insert(root, 14);     
temp = root.left.right.right  
  
succ = inOrderSuccessor( root, temp) 
if succ is not None: 
    print "\nInorder Successor of %d is %d " \ 
            %(temp.data , succ.data) 
else: 
    print "\nInorder Successor doesn't exist"
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

Output of the above program:
Inorder Successor of 14 is 20

Time Complexity: O(h) where h is height of tree.

Method 2 (Search from root)
Parent pointer is NOT needed in this algorithm. The Algorithm is divided into two cases on the basis of right subtree of the input node being empty or not.

Input: node, root // node is the node whose Inorder successor is needed.
output: succ // succ is Inorder successor of node.

1) If right subtree of node is not NULL, then succ lies in right subtree. Do following.
Go to right subtree and return the node with minimum key value in right subtree.
2) If right sbtree of node is NULL, then start from root and us search like technique. Do following.
Travel down the tree, if a node’s data is greater than root’s data then go right side, otherwise go to left side.

struct node * inOrderSuccessor(struct node *root, struct node *n) 
{ 
    // step 1 of the above algorithm 
    if( n->right != NULL ) 
        return minValue(n->right); 
  
    struct node *succ = NULL; 
  
    // Start from root and search for successor down the tree 
    while (root != NULL) 
    { 
        if (n->data < root->data) 
        { 
            succ = root; 
            root = root->left; 
        } 
        else if (n->data > root->data) 
            root = root->right; 
        else
           break; 
    } 
  
    return succ; 
} 

Thanks to R.Srinivasan for suggesting this method.

Time Complexity: O(h) where h is height of tree.

References:
http://net.pku.edu.cn/~course/cs101/2007/resource/Intro2Algorithm/book6/chap13.htm

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

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

C

Java

Python

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