Inorder Successor in Binary Search Tree Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice In the Binary Tree, the Inorder successor of a node is the next node in the Inorder traversal of the Binary Tree. Inorder Successor is NULL for the last node in Inorder traversal. Example:In the below diagram, inorder successor of 8 is 10, inorder successor of 10 is 12 and inorder successor of 14 is 20. We have discussed different methods for Inorder successor in a Binary Tree. These methods either work in O(n) time or expect parent pointer/reference to be present in every node. We can use Binary Search Tree properties to efficiently find the successor in O(h) time. Approach:We follow the idea of normal BST Search. In BST search, we get closer to the key by comparing with the current node. So the last greater key visited during search is the successor. The following cases arise during the search.If we reach null, then the given target does not exist, we return nullIf current node matches the target and right child is not empty, then successor is leftmost node in right subtree.If current node is greater , then it is a potential successor, we mark it as successor and proceed to leftIf current node is smaller or equal to the target, we proceed to right.Below is the implementation of the above approach: C++ // C++ Program to find Inorder Successor in // Binary Search Tree #include <bits/stdc++.h> using namespace std; class Node { public: int data; Node* left; Node* right; Node(int x) { data = x; left = nullptr; right = nullptr; } }; Node* leftMost(Node* node) { Node* curr = node; while (curr->left != nullptr) curr = curr->left; return curr; } Node* getSucc(Node* root, int target) { // Base Case 1: No Inorder Successor if (root == nullptr) return nullptr; // Base Case 2: root is same as target and // right child is not empty if (root->data == target && root->right != nullptr) return leftMost(root->right); // Use BST properties to search for // target and successor Node* succ = nullptr; Node* curr = root; while (curr != nullptr) { // If curr node is greater, then it // is a potential successor if (target < curr->data) { succ = curr; curr = curr->left; } // If smaller, then successor must // be in the right child else if (target >= curr->data) curr = curr->right; } return succ; } int main() { // Construct a BST // 20 // / \ // 8 22 // / \ // 4 12 // / \ // 10 14 Node *root = new Node(20); root->left = new Node(8); root->right = new Node(22); root->left->left = new Node(4); root->left->right = new Node(12); root->left->right->left = new Node(10); root->left->right->right = new Node(14); int target = 14; Node* succ = getSucc(root, target); if (succ != nullptr) cout << succ->data; else cout << "null"; return 0; } C // C Program to find Inorder Successor // in Binary Search Tree #include <stdio.h> #include <stdlib.h> struct Node { int data; struct Node* left; struct Node* right; }; // Function to find leftmost node in // subtree with given root. struct Node* leftMost(struct Node* node) { struct Node* curr = node; while (curr->left != NULL) curr = curr->left; return curr; } struct Node* getSucc(struct Node* root, int target) { // Base Case 1: No Inorder Successor if (root == NULL) return NULL; // Base Case 2: root is same as target and // right child is not empty if (root->data == target && root->right != NULL) return leftMost(root->right); // Use BST properties to search for // target and successor struct Node* succ = NULL; struct Node* curr = root; while (curr != NULL) { // If curr node is greater, then it // is a potential successor if (target < curr->data) { succ = curr; curr = curr->left; } // If smaller, then successor must // be in the right child else if (target >= curr->data) curr = curr->right; } return succ; } struct Node* createNode(int x) { struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->data = x; newNode->left = NULL; newNode->right = NULL; return newNode; } int main() { // Construct a BST // 20 // / \ // 8 22 // / \ // 4 12 // / \ // 10 14 struct Node *root = createNode(20); root->left = createNode(8); root->right = createNode(22); root->left->left = createNode(4); root->left->right = createNode(12); root->left->right->left = createNode(10); root->left->right->right = createNode(14); int target = 14; struct Node* succ = getSucc(root, target); if (succ != NULL) printf("%d", succ->data); else printf("null"); return 0; } Java // Java Program to find Inorder Successor // in Binary Search Tree class Node { int data; Node left, right; Node(int x) { data = x; left = null; right = null; } } class GfG { // Function to find leftmost node in // subtree with given root. static Node leftMost(Node node) { Node curr = node; while (curr.left != null) curr = curr.left; return curr; } static Node getSucc(Node root, int target) { // Base Case 1: No Inorder Successor if (root == null) return null; // Base Case 2: root is same as target and // right child is not empty if (root.data == target && root.right != null) return leftMost(root.right); // Use BST properties to search for // target and successor Node succ = null; Node curr = root; while (curr != null) { // If curr node is greater, then it // is a potential successor if (target < curr.data) { succ = curr; curr = curr.left; } // If smaller, then successor must // be in the right child else if (target >= curr.data) curr = curr.right; } return succ; } public static void main(String[] args) { // Construct a BST // 20 // / \ // 8 22 // / \ // 4 12 // / \ // 10 14 Node root = new Node(20); root.left = new Node(8); root.right = new Node(22); root.left.left = new Node(4); root.left.right = new Node(12); root.left.right.left = new Node(10); root.left.right.right = new Node(14); int target = 14; Node succ = getSucc(root, target); if (succ != null) System.out.println(succ.data); else System.out.println("null"); } } Python # Python Program to find Inorder # successor in Binary Search Tree class Node: def __init__(self, x): self.data = x self.left = None self.right = None # Function to find leftmost node in # subtree with given root. def leftMost(node): curr = node while curr.left is not None: curr = curr.left return curr def getSucc(root, target): # Base Case 1: No Inorder Successor if root is None: return None # Base Case 2: root is same as target and # right child is not empty if root.data == target and root.right is not None: return leftMost(root.right) # Use BST properties to search for # target and successor succ = None curr = root while curr is not None: # If curr node is greater, then it # is a potential successor if target < curr.data: succ = curr curr = curr.left # If smaller, then successor must # be in the right child elif target >= curr.data: curr = curr.right return succ if __name__ == '__main__': # Construct a BST # 20 # / \ # 8 22 # / \ # 4 12 # / \ # 10 14 root = Node(20) root.left = Node(8) root.right = Node(22) root.left.left = Node(4) root.left.right = Node(12) root.left.right.left = Node(10) root.left.right.right = Node(14) target = 14 succ = getSucc(root, target) if succ is not None: print(succ.data) else: print("null") C# // C# Program to find Inorder Successor // in Binary Search Tree using System; class Node { public int data; public Node left, right; public Node(int x) { data = x; left = null; right = null; } } class GfG { // Function to find leftmost node in // subtree with given root. static Node leftMost(Node node) { Node curr = node; while (curr.left != null) curr = curr.left; return curr; } static Node getSucc(Node root, int target) { // Base Case 1: No Inorder Successor if (root == null) return null; // Base Case 2: root is same as target and // right child is not empty if (root.data == target && root.right != null) return leftMost(root.right); // Use BST properties to search for // target and successor Node succ = null; Node curr = root; while (curr != null) { // If curr node is greater, then it // is a potential successor if (target < curr.data) { succ = curr; curr = curr.left; } // If smaller, then successor must // be in the right child else if (target >= curr.data) curr = curr.right; } return succ; } static void Main(string[] args) { // Construct a BST // 20 // / \ // 8 22 // / \ // 4 12 // / \ // 10 14 Node root = new Node(20); root.left = new Node(8); root.right = new Node(22); root.left.left = new Node(4); root.left.right = new Node(12); root.left.right.left = new Node(10); root.left.right.right = new Node(14); int target = 14; Node succ = getSucc(root, target); if (succ != null) Console.WriteLine(succ.data); else Console.WriteLine("null"); } } JavaScript // JavaScript Program to find Inorder // Successor in Binary Search Tree class Node { constructor(x) { this.data = x; this.left = null; this.right = null; } } // Function to find leftmost node in // subtree with given root. function leftMost(node) { let curr = node; while (curr.left !== null) { curr = curr.left; } return curr; } function getSucc(root, target) { // Base Case 1: No Inorder Successor if (root === null) return null; // Base Case 2: root is same as target and // right child is not empty if (root.data === target && root.right !== null) return leftMost(root.right); // Use BST properties to search for // target and successor let succ = null; let curr = root; while (curr !== null) { // If curr node is greater, then it // is a potential successor if (target < curr.data) { succ = curr; curr = curr.left; } // If smaller, then successor must // be in the right child else if (target >= curr.data) { curr = curr.right; } } return succ; } // Construct a BST // 20 // / \ // 8 22 // / \ // 4 12 // / \ // 10 14 const root = new Node(20); root.left = new Node(8); root.right = new Node(22); root.left.left = new Node(4); root.left.right = new Node(12); root.left.right.left = new Node(10); root.left.right.right = new Node(14); const target = 14; const succ = getSucc(root, target); if (succ !== null) console.log(succ.data); else console.log("null"); Output20Time Complexity: O(h), where h is the height of the tree. 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