Insert a node in Binary Search Tree Iteratively
A recursive approach to insert a new node in a BST is already discussed in the post: Binary Search Tree | SET 1. In this post, an iterative approach to insert a node in BST is discussed.
Insertion of a Key
A new key is always inserted at the leaf node. Start searching a key from root till we hit a leaf node. Once a leaf node is found, the new node is added as a child of the leaf node.
Example:
Input:To the given BST insert 40
Output:
Explanation:The new node 40 is a leaf node. Start searching from the root till a leaf node is hit, i.e while searching if a new value is greater than current node move to right child else to left child.
Input:To the given BST insert 600
Output:
Explanation:The new node 600 is a leaf node. Start searching from the root till a leaf node is hit, i.e while searching if a new value is greater than current node move to right child else to left child.
Solution:
Approach:
- It is to be noted that new keys are always inserted at the leaf node.
- Start from root and run a loop until a null pointer is reached.
- Keep the previous pointer of the current node stored.
- If the current node is null then create and insert the new node there and make it as one of the children of the parent/previous node depending on its value.
- If the value of current node is less than the new value then move to the right child of current node else move to the left child.
Below is the implementation of the above approach:
C++
// C++ program to demonstrate insert operation// in binary search tree#include <bits/stdc++.h>using namespace std;// BST nodestruct Node { int key; struct Node *left, *right;};// Utitlity function to create a new nodeNode* newNode(int data){ Node* temp = new Node; temp->key = data; temp->left = NULL; temp->right = NULL; return temp;}// A utility function to insert a new// Node with given key in BSTNode* insert(Node* root, int key){ // Create a new Node containing // the new element Node* newnode = newNode(key); // Pointer to start traversing from root and // traverses downward path to search // where the new node to be inserted Node* x = root; // Pointer y maintains the trailing // pointer of x Node* y = NULL; while (x != NULL) { y = x; if (key < x->key) x = x->left; else x = x->right; } // If the root is NULL i.e the tree is empty // The new node is the root node if (y == NULL) y = newnode; // If the new key is less then the leaf node key // Assign the new node to be its left child else if (key < y->key) y->left = newnode; // else assign the new node its right child else y->right = newnode; // Returns the pointer where the // new node is inserted return y;}// A utility function to do inorder// traversal of BSTvoid Inorder(Node* root){ if (root == NULL) return; else { Inorder(root->left); cout << root->key << " "; Inorder(root->right); }}// Driver codeint main(){ /* 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); // Print inoder traversal of the BST Inorder(root); return 0;} |
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Java
// Java program to demonstrate insert operation // in binary search tree import java.util.*;class solution{ // BST node static class Node { int key; Node left, right; }; // Utitlity function to create a new node static Node newNode(int data) { Node temp = new Node(); temp.key = data; temp.left = null; temp.right = null; return temp; } // A utility function to insert a new // Node with given key in BST static Node insert(Node root, int key) { // Create a new Node containing // the new element Node newnode = newNode(key); // Pointer to start traversing from root and // traverses downward path to search // where the new node to be inserted Node x = root; // Pointer y maintains the trailing // pointer of x Node y = null; while (x != null) { y = x; if (key < x.key) x = x.left; else x = x.right; } // If the root is null i.e the tree is empty // The new node is the root node if (y == null) y = newnode; // If the new key is less then the leaf node key // Assign the new node to be its left child else if (key < y.key) y.left = newnode; // else assign the new node its right child else y.right = newnode; // Returns the pointer where the // new node is inserted return y; } // A utility function to do inorder // traversal of BST static void Inorder(Node root) { if (root == null) return; else { Inorder(root.left); System.out.print( root.key +" "); Inorder(root.right); } } // Driver code public static void main(String args[]) { /* 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); // Print inoder traversal of the BST Inorder(root); } }//contributed by Arnab Kundu |
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Python3
"""Python3 program to demonstrate insert operation in binary search tree """# A Binary Tree Node # Utility function to create a# new tree node class newNode: # Constructor to create a newNode def __init__(self, data): self.key= data self.left = None self.right = self.parent = None# A utility function to insert a new # Node with given key in BST def insert(root, key): # Create a new Node containing # the new element newnode = newNode(key) # Pointer to start traversing from root # and traverses downward path to search # where the new node to be inserted x = root # Pointer y maintains the trailing # pointer of x y = None while (x != None): y = x if (key < x.key): x = x.left else: x = x.right # If the root is None i.e the tree is # empty. The new node is the root node if (y == None): y = newnode # If the new key is less then the leaf node key # Assign the new node to be its left child elif (key < y.key): y.left = newnode # else assign the new node its # right child else: y.right = newnode # Returns the pointer where the # new node is inserted return y # A utility function to do inorder # traversal of BST def Inorder(root) : if (root == None) : return else: Inorder(root.left) print( root.key, end = " " ) Inorder(root.right) # Driver Codeif __name__ == '__main__': """ 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) # Prinoder traversal of the BST Inorder(root)# This code is contributed by# SHUBHAMSINGH10 |
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C#
// C# program to demonstrate insert // operation in binary search treeusing System;class GFG { // BST node class Node { public int key; public Node left, right; }; // Utitlity function to create a new node static Node newNode(int data) { Node temp = new Node(); temp.key = data; temp.left = null; temp.right = null; return temp; } // A utility function to insert a new // Node with given key in BST static Node insert(Node root, int key) { // Create a new Node containing // the new element Node newnode = newNode(key); // Pointer to start traversing from root and // traverses downward path to search // where the new node to be inserted Node x = root; // Pointer y maintains the trailing // pointer of x Node y = null; while (x != null) { y = x; if (key < x.key) x = x.left; else x = x.right; } // If the root is null i.e the tree is empty // The new node is the root node if (y == null) y = newnode; // If the new key is less then the leaf node key // Assign the new node to be its left child else if (key < y.key) y.left = newnode; // else assign the new node its right child else y.right = newnode; // Returns the pointer where the // new node is inserted return y; } // A utility function to do inorder // traversal of BST static void Inorder(Node root) { if (root == null) return; else { Inorder(root.left); Console.Write( root.key +" "); Inorder(root.right); } } // Driver code public static void Main(String []args) { /* 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); // Print inoder traversal of the BST Inorder(root); } } // This code is contributed 29AjayKumar |
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Output:
20 30 40 50 60 70 80
Complexity Analysis:
- Time Complexity : O(h), where h is height of binary search tree. In worst case the height is equal to number of nodes.
- Space Complexity: O(1), no extra space is required.
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