Check if value exists in level-order sorted complete binary tree
Given a level-order sorted complete binary tree, the task is to check whether a key exists in it or not. A complete binary tree has every level except possibly the last, completely filled, with all nodes as far left as possible.
Examples:
7
/ \
10 15
/ \ / \
17 20 30 35
/ \ /
40 41 50
Input: Node = 3
Output: No
Input: Node = 7
Output: Yes
Input: Node = 30
Output: YesApproach A simple O(n) solution would be to fully traverse the tree and check for the key value. However, we can leverage the information that the tree is sorted and do better in terms of time complexity.
- Find out the level where the key may exist. Start at the root node, keep going left until a value which is greater than the key value is encountered. The level before this would contain the key, if at all the key existed in the tree. Let us assume this is level l.
- Now, perform binary search on the nodes of l. Unlike the conventional binary search, the nodes of this level cannot be accessed directly. However, the path from the root to every node in this level can be encoded using the binary logic. For example, consider the 3rd level in the sample tree. It can contain up to 23 = 8 nodes. These nodes can be reached from the root node by going left, left, left; or by going left, left, right; and so on. If the left is denoted by 0 and the right by 1 then the possible ways to reach nodes in this level can be encoded as arr = [000, 001, 010, 011, 100, 101, 110, 111].
- However, this array doesn’t need to be created, binary search can be applied by recursively selecting the middle index and simply generating the l-bit gray code of this index (Refer to this article).
- In case of incomplete paths, simply check for the left part of the array. For instance, the encoded path 011 does not correspond to any value in the sample tree. Since the tree is complete, it ensures that there will be no more elements to the right.
- If the key is found return true, else return false.
Below is the implementation of the above approach:
C++14
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
/* Class containing left and right
child of current node and key value*/
class Node
{
public :
int data;
Node *left, *right;
Node(int item)
{
data = item;
left = right = NULL;
}
};
/* Function to locate which level to check
for the existence of key. */
int findLevel(Node *root, int data)
{
// If the key is less than the root,
// it will certainly not exist in the tree
// because tree is level-order sorted
if (data < root->data)
return -1;
// If the key is equal to the root then
// simply return 0 (zero'th level)
if (data == root->data)
return 0;
int cur_level = 0;
while (true)
{
cur_level++;
root = root->left;
// If the key is found in any leftmost element
// then simply return true
// No need for any extra searching
if (root->data == data)
return -2;
// If key lies between the root data and
// the left child's data OR if key is greater
// than root data and there is no level
// underneath it, return the current level
if (root->data < data
&& (root->left == NULL
|| root->left->data > data))
{
break;
}
}
return cur_level;
}
/* Function to traverse a binary
encoded path and return the value
encountered after traversal. */
int traversePath(Node *root,vector<int> path)
{
for (int i = 0; i < path.size(); i++)
{
int direction = path[i];
// Go left
if (direction == 0)
{
// Incomplete path
if (root->left == NULL)
return -1;
root = root->left;
}
// Go right
else
{
// Incomplete path
if (root->right == NULL)
return -1;
root = root->right;
}
}
// Return the data at the node
return root->data;
}
/* Function to generate gray code of
corresponding binary number of integer i */
vector<int> generateGray(int n, int x)
{
// Create new arraylist to store
// the gray code
vector<int> gray ;
int i = 0;
while (x > 0)
{
gray.push_back(x % 2);
x = x / 2;
i++;
}
// Reverse the encoding till here
reverse(gray.begin(),gray.end());
// Leftmost digits are filled with 0
for (int j = 0; j < n - i; j++)
gray.insert(gray.begin(), 0);
return gray;
}
/* Function to search the key in a
particular level of the tree. */
bool binarySearch(Node *root, int start,
int end, int data,
int level)
{
if (end >= start)
{
// Find the middle index
int mid = (start + end) / 2;
// Encode path from root to this index
// in the form of 0s and 1s where
// 0 means LEFT and 1 means RIGHT
vector<int> encoding = generateGray(level, mid);
// Traverse the path in the tree
// and check if the key is found
int element_found = traversePath(root, encoding);
// If path is incomplete
if (element_found == -1)
// Check the left part of the level
return binarySearch(root, start,
mid - 1, data, level);
if (element_found == data)
return true;
// Check the right part of the level
if (element_found < data)
return binarySearch(root, mid + 1,
end, data, level);
// Check the left part of the level
else
return binarySearch(root, start,
mid - 1, data, level);
}
// Key not found in that level
return false;
}
// Function that returns true if the
// key is found in the tree
bool findKey(Node *root, int data)
{
// Find the level where the key may lie
int level = findLevel(root, data);
// If level is -1 then return false
if (level == -1)
return false;
// If level is -2 i.e. key was found in any
// leftmost element then simply return true
if (level == -2)
return true;
// Apply binary search on the elements
// of that level
return binarySearch(root, 0, (int)pow(2, level) -
1, data, level);
}
// Driver code
int main()
{
/* Consider the following level-order sorted tree
5
/ \
8 10
/ \ / \
13 23 25 30
/ \ /
32 40 50
*/
Node* root = new Node(5);
root->left = new Node(8);
root->right = new Node(10);
root->left->left = new Node(13);
root->left->right = new Node(23);
root->right->left = new Node(25);
root->right->right = new Node(30);
root->left->left->left = new Node(32);
root->left->left->right = new Node(40);
root->left->right->left = new Node(50);
// Keys to be searched
int arr[] = { 5, 8, 9 };
int n = sizeof(arr)/sizeof(int);
for (int i = 0; i < n; i++)
{
if (findKey(root, arr[i]))
cout << ("Yes") << endl;
else
cout << ("No") << endl;
}
}
// This code is contributed by Arnab Kundu
Java
// Java implementation of the approach
import java.util.*;
import java.io.*;
/* Class containing left and right
child of current node and key value*/
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG {
/* Function to locate which level to check
for the existence of key. */
public static int findLevel(Node root, int data)
{
// If the key is less than the root,
// it will certainly not exist in the tree
// because tree is level-order sorted
if (data < root.data)
return -1;
// If the key is equal to the root then
// simply return 0 (zero'th level)
if (data == root.data)
return 0;
int cur_level = 0;
while (true) {
cur_level++;
root = root.left;
// If the key is found in any leftmost element
// then simply return true
// No need for any extra searching
if (root.data == data)
return -2;
// If key lies between the root data and
// the left child's data OR if key is greater
// than root data and there is no level
// underneath it, return the current level
if (root.data < data
&& (root.left == null
|| root.left.data > data)) {
break;
}
}
return cur_level;
}
/* Function to traverse a binary
encoded path and return the value
encountered after traversal. */
public static int traversePath(Node root,
ArrayList<Integer> path)
{
for (int i = 0; i < path.size(); i++) {
int direction = path.get(i);
// Go left
if (direction == 0) {
// Incomplete path
if (root.left == null)
return -1;
root = root.left;
}
// Go right
else {
// Incomplete path
if (root.right == null)
return -1;
root = root.right;
}
}
// Return the data at the node
return root.data;
}
/* Function to generate gray code of
corresponding binary number of integer i */
static ArrayList<Integer> generateGray(int n, int x)
{
// Create new arraylist to store
// the gray code
ArrayList<Integer> gray = new ArrayList<Integer>();
int i = 0;
while (x > 0) {
gray.add(x % 2);
x = x / 2;
i++;
}
// Reverse the encoding till here
Collections.reverse(gray);
// Leftmost digits are filled with 0
for (int j = 0; j < n - i; j++)
gray.add(0, 0);
return gray;
}
/* Function to search the key in a
particular level of the tree. */
public static boolean binarySearch(Node root,
int start,
int end,
int data,
int level)
{
if (end >= start) {
// Find the middle index
int mid = (start + end) / 2;
// Encode path from root to this index
// in the form of 0s and 1s where
// 0 means LEFT and 1 means RIGHT
ArrayList<Integer> encoding = generateGray(level, mid);
// Traverse the path in the tree
// and check if the key is found
int element_found = traversePath(root, encoding);
// If path is incomplete
if (element_found == -1)
// Check the left part of the level
return binarySearch(root, start, mid - 1, data, level);
if (element_found == data)
return true;
// Check the right part of the level
if (element_found < data)
return binarySearch(root, mid + 1, end, data, level);
// Check the left part of the level
else
return binarySearch(root, start, mid - 1, data, level);
}
// Key not found in that level
return false;
}
// Function that returns true if the
// key is found in the tree
public static boolean findKey(Node root, int data)
{
// Find the level where the key may lie
int level = findLevel(root, data);
// If level is -1 then return false
if (level == -1)
return false;
// If level is -2 i.e. key was found in any
// leftmost element then simply return true
if (level == -2)
return true;
// Apply binary search on the elements
// of that level
return binarySearch(root, 0, (int)Math.pow(2, level) - 1, data, level);
}
// Driver code
public static void main(String[] args)
{
/* Consider the following level-order sorted tree
5
/ \
8 10
/ \ / \
13 23 25 30
/ \ /
32 40 50
*/
Node root = new Node(5);
root.left = new Node(8);
root.right = new Node(10);
root.left.left = new Node(13);
root.left.right = new Node(23);
root.right.left = new Node(25);
root.right.right = new Node(30);
root.left.left.left = new Node(32);
root.left.left.right = new Node(40);
root.left.right.left = new Node(50);
// Keys to be searched
int arr[] = { 5, 8, 9 };
int n = arr.length;
for (int i = 0; i < n; i++) {
if (findKey(root, arr[i]))
System.out.println("Yes");
else
System.out.println("No");
}
}
}
Python3
# Python3 implementation of the approach
from sys import maxsize
from collections import deque
INT_MIN = -maxsize
# Class containing left and right
# child of current node and key value
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to locate which level to check
# for the existence of key.
def findLevel(root: Node, data: int) -> int:
# If the key is less than the root,
# it will certainly not exist in the tree
# because tree is level-order sorted
if (data < root.data):
return -1
# If the key is equal to the root then
# simply return 0 (zero'th level)
if (data == root.data):
return 0
cur_level = 0
while True:
cur_level += 1
root = root.left
# If the key is found in any leftmost
# element then simply return true
# No need for any extra searching
if (root.data == data):
return -2
# If key lies between the root data and
# the left child's data OR if key is greater
# than root data and there is no level
# underneath it, return the current level
if (root.data < data and
(root.left == None or
root.left.data > data)):
break
return cur_level
# Function to traverse a binary
# encoded path and return the value
# encountered after traversal.
def traversePath(root: Node, path: list) -> int:
for i in range(len(path)):
direction = path[i]
# Go left
if (direction == 0):
# Incomplete path
if (root.left == None):
return -1
root = root.left
# Go right
else:
# Incomplete path
if (root.right == None):
return -1
root = root.right
# Return the data at the node
return root.data
# Function to generate gray code of
# corresponding binary number of integer i
def generateGray(n: int, x: int) -> list:
# Create new arraylist to store
# the gray code
gray = []
i = 0
while (x > 0):
gray.append(x % 2)
x = x / 2
i += 1
# Reverse the encoding till here
gray.reverse()
# Leftmost digits are filled with 0
for j in range(n - i):
gray.insert(0, gray[0])
return gray
# Function to search the key in a
# particular level of the tree.
def binarySearch(root: Node, start: int,
end: int, data: int,
level: int) -> bool:
if (end >= start):
# Find the middle index
mid = (start + end) / 2
# Encode path from root to this index
# in the form of 0s and 1s where
# 0 means LEFT and 1 means RIGHT
encoding = generateGray(level, mid)
# Traverse the path in the tree
# and check if the key is found
element_found = traversePath(root, encoding)
# If path is incomplete
if (element_found == -1):
# Check the left part of the level
return binarySearch(root, start,
mid - 1, data,
level)
if (element_found == data):
return True
# Check the right part of the level
if (element_found < data):
return binarySearch(root, mid + 1,
end, data, level)
# Check the left part of the level
else:
return binarySearch(root, start,
mid - 1, data,
level)
# Key not found in that level
return False
# Function that returns true if the
# key is found in the tree
def findKey(root: Node, data: int) -> bool:
# Find the level where the key may lie
level = findLevel(root, data)
# If level is -1 then return false
if (level == -1):
return False
# If level is -2 i.e. key was found in any
# leftmost element then simply return true
if (level == -2):
return True
# Apply binary search on the elements
# of that level
return binarySearch(root, 0,
int(pow(2, level) - 1),
data, level)
# Driver code
if __name__ == "__main__":
# Consider the following level
# order sorted tree
'''
5
/ \
8 10
/ \ / \
13 23 25 30
/ \ /
32 40 50
'''
root = Node(5)
root.left = Node(8)
root.right = Node(10)
root.left.left = Node(13)
root.left.right = Node(23)
root.right.left = Node(25)
root.right.right = Node(30)
root.left.left.left = Node(32)
root.left.left.right = Node(40)
root.left.right.left = Node(50)
# Keys to be searched
arr = [ 5, 8, 9 ]
n = len(arr)
for i in range(n):
if (findKey(root, arr[i])):
print("Yes")
else:
print("No")
# This code is contributed by sanjeev2552C#
// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
/* Class containing left and right
child of current node and key value*/
class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG{
/* Function to locate which level to check
for the existence of key. */
public static int findLevel(Node root, int data)
{
// If the key is less than the root,
// it will certainly not exist in the tree
// because tree is level-order sorted
if (data < root.data)
return -1;
// If the key is equal to the root then
// simply return 0 (zero'th level)
if (data == root.data)
return 0;
int cur_level = 0;
while (true)
{
cur_level++;
root = root.left;
// If the key is found in any leftmost element
// then simply return true
// No need for any extra searching
if (root.data == data)
return -2;
// If key lies between the root data and
// the left child's data OR if key is greater
// than root data and there is no level
// underneath it, return the current level
if (root.data < data && (root.left == null ||
root.left.data > data))
{
break;
}
}
return cur_level;
}
/* Function to traverse a binary
encoded path and return the value
encountered after traversal. */
public static int traversePath(Node root, List<int> path)
{
for(int i = 0; i < path.Count; i++)
{
int direction = path[i];
// Go left
if (direction == 0)
{
// Incomplete path
if (root.left == null)
return -1;
root = root.left;
}
// Go right
else
{
// Incomplete path
if (root.right == null)
return -1;
root = root.right;
}
}
// Return the data at the node
return root.data;
}
/* Function to generate gray code of
corresponding binary number of integer i */
static List<int> generateGray(int n, int x)
{
// Create new arraylist to store
// the gray code
List<int> gray = new List<int>();
int i = 0;
while (x > 0)
{
gray.Add(x % 2);
x = x / 2;
i++;
}
// Reverse the encoding till here
gray.Reverse();
// Leftmost digits are filled with 0
for(int j = 0; j < n - i; j++)
gray.Insert(0, 0);
return gray;
}
/* Function to search the key in a
particular level of the tree. */
public static bool binarySearch(Node root, int start,
int end, int data,
int level)
{
if (end >= start)
{
// Find the middle index
int mid = (start + end) / 2;
// Encode path from root to this index
// in the form of 0s and 1s where
// 0 means LEFT and 1 means RIGHT
List<int> encoding = generateGray(level, mid);
// Traverse the path in the tree
// and check if the key is found
int element_found = traversePath(root, encoding);
// If path is incomplete
if (element_found == -1)
// Check the left part of the level
return binarySearch(root, start, mid - 1,
data, level);
if (element_found == data)
return true;
// Check the right part of the level
if (element_found < data)
return binarySearch(root, mid + 1, end,
data, level);
// Check the left part of the level
else
return binarySearch(root, start, mid - 1,
data, level);
}
// Key not found in that level
return false;
}
// Function that returns true if the
// key is found in the tree
public static bool findKey(Node root, int data)
{
// Find the level where the key may lie
int level = findLevel(root, data);
// If level is -1 then return false
if (level == -1)
return false;
// If level is -2 i.e. key was found in any
// leftmost element then simply return true
if (level == -2)
return true;
// Apply binary search on the elements
// of that level
return binarySearch(root, 0, (int)Math.Pow(2, level) - 1,
data, level);
}
// Driver code
static void Main()
{
/* Consider the following level-order sorted tree
5
/ \
8 10
/ \ / \
13 23 25 30
/ \ /
32 40 50
*/
Node root = new Node(5);
root.left = new Node(8);
root.right = new Node(10);
root.left.left = new Node(13);
root.left.right = new Node(23);
root.right.left = new Node(25);
root.right.right = new Node(30);
root.left.left.left = new Node(32);
root.left.left.right = new Node(40);
root.left.right.left = new Node(50);
// Keys to be searched
int []arr = { 5, 8, 9 };
int n = arr.Length;
for(int i = 0; i < n; i++)
{
if (findKey(root, arr[i]))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
}
// This code is contributed by SoumikMondalJavascript
<script>
// JavaScript implementation of the approach
/* Class containing left and right
child of current node and key value*/
class Node {
constructor(item) {
this.left = null;
this.right = null;
this.data = item;
}
}
/* Function to locate which level to check
for the existence of key. */
function findLevel(root, data)
{
// If the key is less than the root,
// it will certainly not exist in the tree
// because tree is level-order sorted
if (data < root.data)
return -1;
// If the key is equal to the root then
// simply return 0 (zero'th level)
if (data == root.data)
return 0;
let cur_level = 0;
while (true) {
cur_level++;
root = root.left;
// If the key is found in any leftmost element
// then simply return true
// No need for any extra searching
if (root.data == data)
return -2;
// If key lies between the root data and
// the left child's data OR if key is greater
// than root data and there is no level
// underneath it, return the current level
if (root.data < data
&& (root.left == null
|| root.left.data > data)) {
break;
}
}
return cur_level;
}
/* Function to traverse a binary
encoded path and return the value
encountered after traversal. */
function traversePath(root, path)
{
for (let i = 0; i < path.length; i++) {
let direction = path[i];
// Go left
if (direction == 0) {
// Incomplete path
if (root.left == null)
return -1;
root = root.left;
}
// Go right
else {
// Incomplete path
if (root.right == null)
return -1;
root = root.right;
}
}
// Return the data at the node
return root.data;
}
/* Function to generate gray code of
corresponding binary number of integer i */
function generateGray(n, x)
{
// Create new arraylist to store
// the gray code
let gray = [];
let i = 0;
while (x > 0) {
gray.push(x % 2);
x = parseInt(x / 2, 10);
i++;
}
// Reverse the encoding till here
gray.reverse();
// Leftmost digits are filled with 0
for (let j = 0; j < n - i; j++)
gray.push(0, 0);
return gray;
}
/* Function to search the key in a
particular level of the tree. */
function binarySearch(root, start, end, data, level)
{
if (end >= start) {
// Find the middle index
let mid = parseInt((start + end) / 2, 10);
// Encode path from root to this index
// in the form of 0s and 1s where
// 0 means LEFT and 1 means RIGHT
let encoding = generateGray(level, mid);
// Traverse the path in the tree
// and check if the key is found
let element_found = traversePath(root, encoding);
// If path is incomplete
if (element_found == -1)
// Check the left part of the level
return binarySearch(root, start, mid - 1, data, level);
if (element_found == data)
return true;
// Check the right part of the level
if (element_found < data)
return binarySearch(root, mid + 1, end, data, level);
// Check the left part of the level
else
return binarySearch(root, start, mid - 1, data, level);
}
// Key not found in that level
return false;
}
// Function that returns true if the
// key is found in the tree
function findKey(root, data)
{
// Find the level where the key may lie
let level = findLevel(root, data);
// If level is -1 then return false
if (level == -1)
return false;
// If level is -2 i.e. key was found in any
// leftmost element then simply return true
if (level == -2)
return true;
// Apply binary search on the elements
// of that level
return binarySearch(root, 0,
Math.pow(2, level) - 1, data, level);
}
/* Consider the following level-order sorted tree
5
/ \
8 10
/ \ / \
13 23 25 30
/ \ /
32 40 50
*/
let root = new Node(5);
root.left = new Node(8);
root.right = new Node(10);
root.left.left = new Node(13);
root.left.right = new Node(23);
root.right.left = new Node(25);
root.right.right = new Node(30);
root.left.left.left = new Node(32);
root.left.left.right = new Node(40);
root.left.right.left = new Node(50);
// Keys to be searched
let arr = [ 5, 8, 9 ];
let n = arr.length;
for (let i = 0; i < n; i++) {
if (findKey(root, arr[i]))
document.write("Yes" + "<br>");
else
document.write("No" + "<br>");
}
</script>Output:
Yes Yes No
Time Complexity: The level can be found in O(logn) time. The time to traverse any path to perform binary search is O(logn). Further, in the worst case, the level may have at most n/2 nodes.
Therefore, the time complexity of performing search becomes O(logn)*O(log(n/2)) = O(logn)^2.
Auxiliary Space: O(1).
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