Inorder predecessor in Binary Search Tree Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a Binary Search Tree, the task is to find the In-Order predecessor of a given target key.In a Binary Search Tree (BST), the Inorder predecessor of a node is the previous node in the Inorder traversal of the BST. The Inorder predecessor is NULL for the first node in the Inorder traversal.Examples:In the below diagram, inorder predecessor of 8 is 4, inorder predecessor of 10 is 8 and inorder predecessor of 14 is 12.We have discussed different methods for Inorder predecessor in 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 lesser key visited during search is the predecessor. Step by step approach:Initialize current node to root and predecessor to NULL.If current node is NULL, return NULL (no predecessor exists).If target > current node value, current node is potential predecessor - update predecessor and move right.If target < current node value, move left.If target equals current node value and left subtree exists, return rightmost node in left subtree.If target equals current node value and no left subtree, break (predecessor already found).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; } }; // Function to find the rightmost (maximum) // node in the left subtree Node* rightmost(Node* node) { while (node->right != nullptr) { node = node->right; } return node; } Node* getPred(Node* root, int target) { // No Inorder predecessor if (root == nullptr) return nullptr; // Use BST properties to search for the // target and predecessor Node* pred = nullptr; Node* curr = root; while (curr != nullptr) { // If the current node's data is less than the target, // it is a potential predecessor if (target > curr->data) { pred = curr; curr = curr->right; } // If greater, move to the left child else if (target < curr->data) { curr = curr->left; } // If equal else { // If the left subtree of target node exists // then predecessor will be right most node // of the left subtree. if (curr->left != nullptr) { return rightmost(curr->left); } // Otherwise, the predecessor has already // been stored in the pred variable break; } } return pred; } 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 = 12; Node* pred = getPred(root, target); if (pred != nullptr) cout << pred->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 the rightmost (maximum) // node in the left subtree struct Node* rightmost(struct Node* node) { while (node->right != NULL) { node = node->right; } return node; } struct Node* getPred(struct Node* root, int target) { // No Inorder predecessor if (root == NULL) return NULL; // Use BST properties to search for the // target and predecessor struct Node* pred = NULL; struct Node* curr = root; while (curr != NULL) { // If the current node's data is less than the target, // it is a potential predecessor if (target > curr->data) { pred = curr; curr = curr->right; } // If greater, move to the left child else if (target < curr->data) { curr = curr->left; } // If equal else { // If the left subtree of target node exists // then predecessor will be right most node // of the left subtree. if (curr->left != NULL) { return rightmost(curr->left); } // Otherwise, the predecessor has already // been stored in the pred variable break; } } return pred; } 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 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 = 12; struct Node* pred = getPred(root, target); if (pred != NULL) printf("%d", pred->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 the rightmost (maximum) // node in the left subtree static Node rightmost(Node node) { while (node.right != null) { node = node.right; } return node; } static Node getPred(Node root, int target) { // No Inorder predecessor if (root == null) return null; // Use BST properties to search for the // target and predecessor Node pred = null; Node curr = root; while (curr != null) { // If the current node's data is less than the target, // it is a potential predecessor if (target > curr.data) { pred = curr; curr = curr.right; } // If greater, move to the left child else if (target < curr.data) { curr = curr.left; } // If equal else { // If the left subtree of target node exists // then predecessor will be right most node // of the left subtree. if (curr.left != null) { return rightmost(curr.left); } // Otherwise, the predecessor has already // been stored in the pred variable break; } } return pred; } public static void main(String[] args) { // Construct a BST 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 = 12; Node pred = getPred(root, target); if (pred != null) System.out.print(pred.data); else System.out.print("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 the rightmost (maximum) # node in the left subtree def rightmost(node): while node.right is not None: node = node.right return node def getPred(root, target): # No Inorder predecessor if root is None: return None # Use BST properties to search for the # target and predecessor pred = None curr = root while curr is not None: # If the current node's data is less than the target, # it is a potential predecessor if target > curr.data: pred = curr curr = curr.right # If greater, move to the left child elif target < curr.data: curr = curr.left # If equal else: # If the left subtree of target node exists # then predecessor will be right most node # of the left subtree. if curr.left is not None: return rightmost(curr.left) # Otherwise, the predecessor has already # been stored in the pred variable break return pred if __name__ == "__main__": # Construct a BST 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 = 12 pred = getPred(root, target) if pred is not None: print(pred.data) else: print("null") C# // C# Program to find Inorder Successor in // Binary Search Tree using System; using System.Collections.Generic; class Node { public int data; public Node left, right; public Node(int x) { data = x; left = null; right = null; } } class GfG { // Function to find the rightmost (maximum) // node in the left subtree static Node rightmost(Node node) { while (node.right != null) { node = node.right; } return node; } static Node getPred(Node root, int target) { // No Inorder predecessor if (root == null) return null; // Use BST properties to search for the // target and predecessor Node pred = null; Node curr = root; while (curr != null) { // If the current node's data is less than the target, // it is a potential predecessor if (target > curr.data) { pred = curr; curr = curr.right; } // If greater, move to the left child else if (target < curr.data) { curr = curr.left; } // If equal else { // If the left subtree of target node exists // then predecessor will be right most node // of the left subtree. if (curr.left != null) { return rightmost(curr.left); } // Otherwise, the predecessor has already // been stored in the pred variable break; } } return pred; } static void 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 = 12; Node pred = getPred(root, target); Console.WriteLine(pred != null ? pred.data.ToString() : "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 the rightmost (maximum) // node in the left subtree function rightmost(node) { while (node.right !== null) { node = node.right; } return node; } function getPred(root, target) { // No Inorder predecessor if (root === null) return null; // Use BST properties to search for the // target and predecessor let pred = null; let curr = root; while (curr !== null) { // If the current node's data is less than the target, // it is a potential predecessor if (target > curr.data) { pred = curr; curr = curr.right; } // If greater, move to the left child else if (target < curr.data) { curr = curr.left; } // If equal else { // If the left subtree of target node exists // then predecessor will be right most node // of the left subtree. if (curr.left !== null) { return rightmost(curr.left); } // Otherwise, the predecessor has already // been stored in the pred variable break; } } return pred; } // Construct a BST // 20 // / \ // 8 22 // / \ // 4 12 // / \ // 10 14 let 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); let target = 12; let pred = getPred(root, target); if (pred !== null) console.log(pred.data); else console.log("null"); Output10Time Complexity: O(h), where h is the height of the tree. 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