Given a Binary Tree and a key to be searched in it, write an iterative method that returns true if key is present in Binary Tree, else false.
For example, in the following tree, if the searched key is 3, then function should return true and if the searched key is 12, then function should return false.
One thing is sure that we need to traverse complete tree to decide whether key is present or not. We can use any of the following traversals to iteratively search a key in a given binary tree.
1) Iterative Level Order Traversal.
2) Iterative Inorder Traversal
3) Iterative Preorder Traversal
4) Iterative Postorder Traversal
Below is iterative Level Order Traversal based solution to search an item x in binary tree.
C++
// Iterative level order traversal // based method to search in Binary Tree #include<bits/stdc++.h> using namespace std; /* A binary tree node has data, left child and right child */class node { public: int data; node* left; node* right; /* Constructor that allocates a new node with the given data and NULL left and right pointers. */ node(int data){ this->data = data; this->left = NULL; this->right = NULL; } }; // An iterative process to search // an element x in a given binary tree bool iterativeSearch(node *root, int x) { // Base Case if (root == NULL) return false; // Create an empty queue for // level order traversal queue<node *> q; // Enqueue Root and initialize height q.push(root); // Queue based level order traversal while (q.empty() == false) { // See if current node is same as x node *node = q.front(); if (node->data == x) return true; // Remove current node and enqueue its children q.pop(); if (node->left != NULL) q.push(node->left); if (node->right != NULL) q.push(node->right); } return false; } // Driver code int main() { node* NewRoot=NULL; node *root = new node(2); root->left = new node(7); root->right = new node(5); root->left->right = new node(6); root->left->right->left=new node(1); root->left->right->right=new node(11); root->right->right=new node(9); root->right->right->left=new node(4); iterativeSearch(root, 6)? cout << "Found\n": cout << "Not Found\n"; iterativeSearch(root, 12)? cout << "Found\n": cout << "Not Found\n"; return 0; } // This code is contributed by rathbhupendra |
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C
// Iterative level order traversal based method to search in Binary Tree #include <iostream> #include <queue> using namespace std; /* A binary tree node has data, left child and right child */struct node { int data; struct node* left, *right; }; /* 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 = new struct node; node->data = data; node->left = node->right = NULL; return(node); } // An iterative process to search an element x in a given binary tree bool iterativeSearch(node *root, int x) { // Base Case if (root == NULL) return false; // Create an empty queue for level order traversal queue<node *> q; // Enqueue Root and initialize height q.push(root); // Queue based level order traversal while (q.empty() == false) { // See if current node is same as x node *node = q.front(); if (node->data == x) return true; // Remove current node and enqueue its children q.pop(); if (node->left != NULL) q.push(node->left); if (node->right != NULL) q.push(node->right); } return false; } // Driver program int main(void) { struct node*NewRoot=NULL; struct node *root = newNode(2); root->left = newNode(7); root->right = newNode(5); root->left->right = newNode(6); root->left->right->left=newNode(1); root->left->right->right=newNode(11); root->right->right=newNode(9); root->right->right->left=newNode(4); iterativeSearch(root, 6)? cout << "Found\n": cout << "Not Found\n"; iterativeSearch(root, 12)? cout << "Found\n": cout << "Not Found\n"; return 0; } |
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Java
// Iterative level order traversal // based method to search in Binary Tree import java.util.*; class GFG { /* A binary tree node has data, left child and right child */static class node { int data; node left; node right; /* Constructor that allocates a new node with the given data and null left and right pointers. */ node(int data) { this.data = data; this.left = null; this.right = null; } }; // An iterative process to search // an element x in a given binary tree static boolean iterativeSearch(node root, int x) { // Base Case if (root == null) return false; // Create an empty queue for // level order traversal Queue<node > q = new LinkedList(); // Enqueue Root and initialize height q.add(root); // Queue based level order traversal while (q.size() > 0) { // See if current node is same as x node node = q.peek(); if (node.data == x) return true; // Remove current node and enqueue its children q.remove(); if (node.left != null) q.add(node.left); if (node.right != null) q.add(node.right); } return false; } // Driver code public static void main(String ags[]) { node NewRoot = null; node root = new node(2); root.left = new node(7); root.right = new node(5); root.left.right = new node(6); root.left.right.left = new node(1); root.left.right.right = new node(11); root.right.right = new node(9); root.right.right.left = new node(4); System.out.print((iterativeSearch(root, 6)? "Found\n": "Not Found\n")); System.out.print((iterativeSearch(root, 12)? "Found\n": "Not Found\n")); } } // This code is contributed by Arnab Kundu |
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Python3
# Iterative level order traversal based # method to search in Binary Tree # importing Queue from queue import Queue # Helper function that allocates a # new node with the given data and # None left and right pointers. class newNode: def __init__(self, data): self.data = data self.left = self.right = None # An iterative process to search an # element x in a given binary tree def iterativeSearch(root, x): # Base Case if (root == None): return False # Create an empty queue for level # order traversal q = Queue() # Enqueue Root and initialize height q.put(root) # Queue based level order traversal while (q.empty() == False): # See if current node is same as x node = q.queue[0] if (node.data == x): return True # Remove current node and # enqueue its children q.get() if (node.left != None): q.put(node.left) if (node.right != None): q.put(node.right) return False # Driver Code if __name__ == '__main__': root = newNode(2) root.left = newNode(7) root.right = newNode(5) root.left.right = newNode(6) root.left.right.left = newNode(1) root.left.right.right = newNode(11) root.right.right = newNode(9) root.right.right.left = newNode(4) if iterativeSearch(root, 6): print("Found") else: print("Not Found") if iterativeSearch(root, 12): print("Found") else: print("Not Found") # This code is contributed by PranchalK |
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C#
// Iterative level order traversal // based method to search in Binary Tree using System; using System.Collections.Generic; class GFG { /* A binary tree node has data, left child and right child */public class node { public int data; public node left; public node right; /* Constructor that allocates a new node with the given data and null left and right pointers. */ public node(int data) { this.data = data; this.left = null; this.right = null; } }; // An iterative process to search // an element x in a given binary tree static Boolean iterativeSearch(node root, int x) { // Base Case if (root == null) return false; // Create an empty queue for // level order traversal Queue<node > q = new Queue<node>(); // Enqueue Root and initialize height q.Enqueue(root); // Queue based level order traversal while (q.Count > 0) { // See if current node is same as x node node = q.Peek(); if (node.data == x) return true; // Remove current node and // enqueue its children q.Dequeue(); if (node.left != null) q.Enqueue(node.left); if (node.right != null) q.Enqueue(node.right); } return false; } // Driver code public static void Main(String []ags) { node root = new node(2); root.left = new node(7); root.right = new node(5); root.left.right = new node(6); root.left.right.left = new node(1); root.left.right.right = new node(11); root.right.right = new node(9); root.right.right.left = new node(4); Console.WriteLine((iterativeSearch(root, 6) ? "Found\n" : "Not Found")); Console.Write((iterativeSearch(root, 12) ? "Found\n" : "Not Found\n")); } } // This code is contributed by Rajput-Ji |
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Output:
Found Not Found
Below implementation uses Iterative Preorder Traversal to find x in Binary Tree
C/C++
// An iterative method to search an item in Binary Tree #include <iostream> #include <stack> using namespace std; /* A binary tree node has data, left child and right child */struct node { int data; struct node* left, *right; }; /* 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 = new struct node; node->data = data; node->left = node->right = NULL; return(node); } // iterative process to search an element x in a given binary tree bool iterativeSearch(node *root, int x) { // Base Case if (root == NULL) return false; // Create an empty stack and push root to it stack<node *> nodeStack; nodeStack.push(root); // Do iterative preorder traversal to search x while (nodeStack.empty() == false) { // See the top item from stack and check if it is same as x struct node *node = nodeStack.top(); if (node->data == x) return true; nodeStack.pop(); // Push right and left children of the popped node to stack if (node->right) nodeStack.push(node->right); if (node->left) nodeStack.push(node->left); } return false; } // Driver program int main(void) { struct node*NewRoot=NULL; struct node *root = newNode(2); root->left = newNode(7); root->right = newNode(5); root->left->right = newNode(6); root->left->right->left=newNode(1); root->left->right->right=newNode(11); root->right->right=newNode(9); root->right->right->left=newNode(4); iterativeSearch(root, 6)? cout << "Found\n": cout << "Not Found\n"; iterativeSearch(root, 12)? cout << "Found\n": cout << "Not Found\n"; return 0; } |
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Java
// An iterative method to search an item in Binary Tree import java.util.*; class GFG { /* A binary tree node has data, left child and right child */static class node { int data; node left, right; }; /* Helper function that allocates a new node with the given data and null left and right pointers.*/static node newNode(int data) { node node = new node(); node.data = data; node.left = node.right = null; return(node); } // iterative process to search // an element x in a given binary tree static boolean iterativeSearch(node root, int x) { // Base Case if (root == null) return false; // Create an empty stack and push root to it Stack<node> nodeStack = new Stack<node>(); nodeStack.push(root); // Do iterative preorder traversal to search x while (nodeStack.empty() == false) { // See the top item from stack and // check if it is same as x node node = nodeStack.peek(); if (node.data == x) return true; nodeStack.pop(); // Push right and left children // of the popped node to stack if (node.right != null) nodeStack.push(node.right); if (node.left != null) nodeStack.push(node.left); } return false; } // Driver Code public static void main(String[] args) { node NewRoot = null; node root = newNode(2); root.left = newNode(7); root.right = newNode(5); root.left.right = newNode(6); root.left.right.left = newNode(1); root.left.right.right = newNode(11); root.right.right = newNode(9); root.right.right.left = newNode(4); if(iterativeSearch(root, 6)) System.out.println("Found"); else System.out.println("Not Found"); if(iterativeSearch(root, 12)) System.out.println("Found"); else System.out.println("Not Found"); } } // This code is contributed by 29AjayKumar |
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Python3
# An iterative Python3 code to search # an item in Binary Tree ''' A binary tree node has data, left child and right child '''class newNode: # Construct to create a newNode def __init__(self, key): self.data = key self.left = None self.right = None # iterative process to search an element x # in a given binary tree def iterativeSearch(root,x): # Base Case if (root == None): return False # Create an empty stack and # append root to it nodeStack = [] nodeStack.append(root) # Do iterative preorder traversal to search x while (len(nodeStack)): # See the top item from stack and # check if it is same as x node = nodeStack[0] if (node.data == x): return True nodeStack.pop(0) # append right and left children # of the popped node to stack if (node.right): nodeStack.append(node.right) if (node.left): nodeStack.append(node.left) return False # Driver Code root = newNode(2) root.left = newNode(7) root.right = newNode(5) root.left.right = newNode(6) root.left.right.left = newNode(1) root.left.right.right = newNode(11) root.right.right = newNode(9) root.right.right.left = newNode(4) if iterativeSearch(root, 6): print("Found") else: print("Not Found") if iterativeSearch(root, 12): print("Found") else: print("Not Found") # This code is contributed by SHUBHAMSINGH10 |
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C#
// An iterative method to search an item in Binary Tree using System; using System.Collections.Generic; class GFG { /* A binary tree node has data, left child and right child */class node { public int data; public node left, right; }; /* Helper function that allocates a new node with the given data and null left and right pointers.*/static node newNode(int data) { node node = new node(); node.data = data; node.left = node.right = null; return(node); } // iterative process to search // an element x in a given binary tree static bool iterativeSearch(node root, int x) { // Base Case if (root == null) return false; // Create an empty stack and.Push root to it Stack<node> nodeStack = new Stack<node>(); nodeStack.Push(root); // Do iterative preorder traversal to search x while (nodeStack.Count != 0) { // See the top item from stack and // check if it is same as x node node = nodeStack.Peek(); if (node.data == x) return true; nodeStack.Pop(); // Push right and left children // of the.Popped node to stack if (node.right != null) nodeStack.Push(node.right); if (node.left != null) nodeStack.Push(node.left); } return false; } // Driver Code public static void Main(String[] args) { node root = newNode(2); root.left = newNode(7); root.right = newNode(5); root.left.right = newNode(6); root.left.right.left = newNode(1); root.left.right.right = newNode(11); root.right.right = newNode(9); root.right.right.left = newNode(4); if(iterativeSearch(root, 6)) Console.WriteLine("Found"); else Console.WriteLine("Not Found"); if(iterativeSearch(root, 12)) Console.WriteLine("Found"); else Console.WriteLine("Not Found"); } } // This code is contributed by PrinciRaj1992 |
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Output:
Found Not Found
Similarly, Iterative Inorder and Iterative Postorder traversals can be used.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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