Inorder predecessor and successor for a given key in BST | Iterative Approach

Given a BST and a key. The task is to find the inorder successor and predecessor of the given key. In case, if either of predecessor or successor is not present, then print -1.

Examples:

Input:          50
               /  \
              /    \
            30     70
           / \     / \
          /   \   /   \
         20   40 60   80
            key = 65
Output: Predecessor : 60
        Successor : 70

Input:          50
               /  \
              /    \
            30     70
           / \     / \
          /   \   /   \
         20   40 60   80
            key = 100
Output: predecessor : 80
        successor : -1
Explanation: As no node in BST has key value greater 
than 100 so -1 is printed for successor.

In the previous post, a recursive solution has been discussed. The problem can be solved using an iterative approach. To solve the problem, the three cases while searching for the key has to be dealt with which are as described below:

  1. Root is the given key: In this case, if the left subtree is not NULL, then predecessor is the rightmost node in left subtree and if right subtree is not NULL, then successor is the leftmost node in right subtree.
  2. Root is greater than key: In this case, the key is present in left subtree of root. So search for the key in left subtree by setting root = root->left. Note that root could be an inorder successor of given key. In case the key has no right subtree, the root will be its successor.
  3. Root is less than key: In this case, key is present in right subtree of root. So search for the key in right subtree by setting root = root->right. Note that root could be an inorder predecessor of given key. In case the key has no left subtree, the root will be its predecessor.

Below is the implementation of above approach:

C++

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// C++ program to find predecessor
// and successor in a BST
#include <bits/stdc++.h>
using namespace std;
  
// BST Node
struct Node {
    int key;
    struct Node *left, *right;
};
  
// Function that finds predecessor and successor of key in BST.
void findPreSuc(Node* root, Node*& pre, Node*& suc, int key)
{
    if (root == NULL)
        return;
  
    // Search for given key in BST.
    while (root != NULL) {
  
        // If root is given key.
        if (root->key == key) {
  
            // the minimum value in right subtree
            // is predecessor.
            if (root->right) {
                suc = root->right;
                while (suc->left)
                    suc = suc->left;
            }
  
            // the maximum value in left subtree
            // is successor.
            if (root->left) {
                pre = root->left;
                while (pre->right)
                    pre = pre->right;
            }
  
            return;
        }
  
        // If key is greater than root, then
        // key lies in right subtree. Root
        // could be predecessor if left
        // subtree of key is null.
        else if (root->key < key) {
            pre = root;
            root = root->right;
        }
  
        // If key is smaller than root, then
        // key lies in left subtree. Root
        // could be successor if right
        // subtree of key is null.
        else {
            suc = root;
            root = root->left;
        }
    }
}
  
// 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 << "-1";
  
    if (suc != NULL)
        cout << "Successor is " << suc->key;
    else
        cout << "-1";
    return 0;
}

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Python3

# Python3 program to find predecessor
# and successor in a BST

# A utility function to create a
# new BST node
class newNode:

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

# Function that finds predecessor and
# successor of key in BST.
def findPreSuc(root, pre, suc, key):
if root == None:
return

# Search for given key in BST.
while root != None:

# If root is given key.
if root.key == key:

# the minimum value in right
# subtree is predecessor.
if root.right:
suc[0] = root.right
while suc[0].left:
suc[0] = suc[0].left

# the maximum value in left
# subtree is successor.
if root.left:
pre[0] = root.left
while pre[0].right:
pre[0] = pre[0].right

return

# If key is greater than root, then
# key lies in right subtree. Root
# could be predecessor if left
# subtree of key is null.
elif root.key < key: pre[0] = root root = root.right # If key is smaller than root, then # key lies in left subtree. Root # could be successor if right # subtree of key is null. else: suc[0] = root root = root.left # A utility function to insert # a new node with given key in BST def insert(node, key): if node == None: return newNode(key) if key < node.key: node.left = insert(node.left, key) else: node.right = insert(node.right, key) return node # Driver Code if __name__ == '__main__': 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) pre, suc = [None], [None] findPreSuc(root, pre, suc, key) if pre[0] != None: print("Predecessor is", pre[0].key) else: print("-1") if suc[0] != None: print("Successor is", suc[0].key) else: print("-1") # This code is contributed by PranchalK [tabbyending]

Output:

Predecessor is 60
Successor is 70

Time Complexity: O(N)
Auxiliary Space: O(1)

Related Article: https://www.geeksforgeeks.org/inorder-predecessor-successor-given-key-bst/



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Improved By : PranchalKatiyar