Subtree with given sum in a Binary Tree
Last Updated :
23 Jul, 2025
You are given a binary tree and a given sum. The task is to check if there exists a subtree whose sum of all nodes is equal to the given sum.
Examples:
Input : key = 11
Output: True
Explanation: sum of all nodes of subtree {2, 4, 5} = 11.
Input : key = 6
Output: False
Explanation: No subtree whose sum of all nodes = 6.
[Expected Approach - 1] Using Recursion - O(n) Time and O(h) Space
The idea is to traverse the binary tree recursively, calculating the sum of the subtree rooted at each node. At each node, we compute the sum of its left and right subtrees and add the node's value itself. If the sum of any subtree matches the given key sum, we mark it as found. We use a helper function subtreeSum() to compute the sum of each subtree and a flag foundSum to track if the key sum is found.
Below is the implementation of the above approach:
C++
// C++ program to find if there is a subtree with
// given sum
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node* left;
Node* right;
Node(int x) {
data = x;
left = nullptr;
right = nullptr;
}
};
// Function to calculate the sum of the subtree rooted at the given node
int subtreeSum(Node* root, int& foundSum, int target) {
// Base case: If the node is NULL, return 0
if (root == nullptr)
return 0;
// Calculate the sum of the current subtree
int currentSum = root->data +
subtreeSum(root->left, foundSum, target) +
subtreeSum(root->right, foundSum, target);
// If the current subtree sum matches the target,
// set foundSum to 1
if (currentSum == target) {
foundSum = 1;
}
// Return the current subtree sum
return currentSum;
}
// Function to check if there is a subtree with the given sum
bool hasSubtreeWithGivenSum(Node* root, int sum) {
int foundSum = 0;
// Traverse the tree and check for the target sum
subtreeSum(root, foundSum, sum);
// Return true if a subtree with the given sum was found
return foundSum == 1;
}
int main() {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
Node* root = new Node(1);
root->left = new Node(3);
root->right = new Node(6);
root->left->left = new Node(5);
root->left->right = new Node(9);
root->right->left = new Node(8);
if(hasSubtreeWithGivenSum(root, 17)) {
cout << "True";
}
else cout << "False";
return 0;
}
// C program to find if there is a subtree with
// given sum
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Function to calculate the sum of the subtree rooted at the given node
int subtreeSum(struct Node* root, int* foundSum, int target) {
// Base case: If the node is NULL, return 0
if (root == NULL)
return 0;
// Calculate the sum of the current subtree
int currentSum = root->data +
subtreeSum(root->left, foundSum, target) +
subtreeSum(root->right, foundSum, target);
// If the current subtree sum matches the target,
// set foundSum to 1
if (currentSum == target) {
*foundSum = 1;
}
// Return the current subtree sum
return currentSum;
}
// Function to check if there is a subtree with the given sum
int hasSubtreeWithGivenSum(struct Node* root, int sum) {
int foundSum = 0;
// Traverse the tree and check for the target sum
subtreeSum(root, &foundSum, sum);
// Return true if a subtree with the given sum was found
return foundSum == 1;
}
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() {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
struct Node* root = createNode(1);
root->left = createNode(3);
root->right = createNode(6);
root->left->left = createNode(5);
root->left->right = createNode(9);
root->right->left = createNode(8);
if(hasSubtreeWithGivenSum(root, 17)) {
printf("True");
}
else printf("False");
return 0;
}
// Java program to find if there is a subtree with
// given sum
class Node {
int data;
Node left, right;
Node(int x) {
data = x;
left = right = null;
}
}
class GfG {
// Function to calculate the sum of the subtree
// rooted at the given node
static int subtreeSum(Node root, boolean[] foundSum, int target) {
// Base case: If the node is NULL, return 0
if (root == null)
return 0;
// Calculate the sum of the current subtree
int currentSum = root.data +
subtreeSum(root.left, foundSum, target) +
subtreeSum(root.right, foundSum, target);
// If the current subtree sum matches the target,
// set foundSum to true
if (currentSum == target) {
foundSum[0] = true;
}
// Return the current subtree sum
return currentSum;
}
// Function to check if there is a subtree with the given sum
static boolean hasSubtreeWithGivenSum(Node root, int sum) {
boolean[] foundSum = new boolean[1];
// Traverse the tree and check for the target sum
subtreeSum(root, foundSum, sum);
// Return true if a subtree with the given sum was found
return foundSum[0];
}
public static void main(String[] args) {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
Node root = new Node(1);
root.left = new Node(3);
root.right = new Node(6);
root.left.left = new Node(5);
root.left.right = new Node(9);
root.right.left = new Node(8);
System.out.println(hasSubtreeWithGivenSum(root, 17));
}
}
# Python program to find if there is a subtree with
# given sum
# Node definition for a binary tree
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
# Function to calculate the sum of the subtree
# rooted at the given node
def subtreeSum(root, foundSum, target):
# Base case: If the node is NULL, return 0
if root is None:
return 0
# Calculate the sum of the current subtree
currentSum = (root.data +
subtreeSum(root.left, foundSum, target) +
subtreeSum(root.right, foundSum, target))
# If the current subtree sum matches the target,
# set foundSum to True
if currentSum == target:
foundSum[0] = True
# Return the current subtree sum
return currentSum
# Function to check if there is a subtree with the given sum
def hasSubtreeWithGivenSum(root, sum):
foundSum = [False]
# Traverse the tree and check for the target sum
subtreeSum(root, foundSum, sum)
# Return true if a subtree with the given sum was found
return foundSum[0]
if __name__ == "__main__":
# Creating the given binary tree
# 1
# / \
# 3 6
# / \ /
# 5 9 8
root = Node(1)
root.left = Node(3)
root.right = Node(6)
root.left.left = Node(5)
root.left.right = Node(9)
root.right.left = Node(8)
print(hasSubtreeWithGivenSum(root, 17))
// C# program to find if there is a subtree with
// given sum
using System;
class Node {
public int data;
public Node left, right;
public Node(int x) {
data = x;
left = right = null;
}
}
class GfG {
// Function to calculate the sum of the subtree
// rooted at the given node
static int SubtreeSum(Node root, ref bool foundSum, int target) {
// Base case: If the node is NULL, return 0
if (root == null)
return 0;
// Calculate the sum of the current subtree
int currentSum = root.data +
SubtreeSum(root.left, ref foundSum, target) +
SubtreeSum(root.right, ref foundSum, target);
// If the current subtree sum matches the target,
// set foundSum to true
if (currentSum == target) {
foundSum = true;
}
// Return the current subtree sum
return currentSum;
}
// Function to check if there is a subtree with the given sum
static bool HasSubtreeWithGivenSum(Node root, int sum) {
bool foundSum = false;
// Traverse the tree and check for the target sum
SubtreeSum(root, ref foundSum, sum);
// Return true if a subtree with the given sum was found
return foundSum;
}
static void Main(string[] args) {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
Node root = new Node(1);
root.left = new Node(3);
root.right = new Node(6);
root.left.left = new Node(5);
root.left.right = new Node(9);
root.right.left = new Node(8);
Console.WriteLine(HasSubtreeWithGivenSum(root, 17));
}
}
// JavaScript program to find if there is a subtree with
// given sum
// Node definition for a binary tree
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
// Function to calculate the sum of the subtree
// rooted at the given node
function subtreeSum(root, foundSum, target) {
// Base case: If the node is NULL, return 0
if (root === null)
return 0;
// Calculate the sum of the current subtree
let currentSum = root.data +
subtreeSum(root.left, foundSum, target) +
subtreeSum(root.right, foundSum, target);
// If the current subtree sum matches the target,
// set foundSum to true
if (currentSum === target) {
foundSum.value = true;
}
// Return the current subtree sum
return currentSum;
}
// Function to check if there is a subtree with the given sum
function hasSubtreeWithGivenSum(root, sum) {
let foundSum = { value: false };
// Traverse the tree and check for the target sum
subtreeSum(root, foundSum, sum);
// Return true if a subtree with the given sum was found
return foundSum.value;
}
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
let root = new Node(1);
root.left = new Node(3);
root.right = new Node(6);
root.left.left = new Node(5);
root.left.right = new Node(9);
root.right.left = new Node(8);
console.log(hasSubtreeWithGivenSum(root, 17));
Time Complexity: O(n), As we are visiting every node once.
Auxiliary space: O(h), here h is the height of the tree and the extra space is used due to the recursion call stack.
[Expected Approach - 2] Using HashMap- O(n) Time and O(n) Space
The idea is to use an iterative depth-first traversal of the binary tree while maintaining a running sum of the node values encountered so far. We will use a hashmap to keep track of the prefix sums encountered during the traversal.
Follow the steps below to solve the problem:
- Start with the root node of the binary tree and push it onto the stack.
- While the stack is not empty, repeatedly:
- Pop the top node from the stack.
- Update the running sum by adding the current node’s value.
- Check if (current sum - target sum) exists in the prefixSumMap. If it does, it means there is a subtree with the required sum.
- Add or update the current running sum in the prefixSumMap to keep track of sums encountered.
- Push the right and left children of the current node onto the stack (if they exist) to continue the traversal.
Below is the implementation of the above approach:
C++
// C++ program to find if there is a subtree with
// given sum
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int val;
Node* left;
Node* right;
Node(int x) {
val = x;
left = nullptr;
right = nullptr;
}
};
// Function to check if there is a subtree with the given sum
bool hasSubtreeWithGivenSum(Node* root, int target) {
if (root == nullptr)
return false;
// Map to store the prefix sums
unordered_map<int, int> prefixSumMap;
prefixSumMap[0] = 1;
int sum = 0;
stack<Node*> st;
st.push(root);
// Iterative depth-first traversal
while (!st.empty()) {
Node* curr = st.top();
st.pop();
// Update the running sum with the
// current node's value
sum += curr->val;
// Check if we have encountered (sum - target) before
if (prefixSumMap.find(sum - target) != prefixSumMap.end()) {
return true;
}
// Add the current running sum to the map
prefixSumMap[sum] = 1;
// Traverse the right and left children
if (curr->right) {
st.push(curr->right);
}
if (curr->left) {
st.push(curr->left);
}
}
// If no subtree with the given sum is found
return false;
}
int main() {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
Node* root = new Node(1);
root->left = new Node(3);
root->right = new Node(6);
root->left->left = new Node(5);
root->left->right = new Node(9);
root->right->left = new Node(8);
if(hasSubtreeWithGivenSum(root, 17)) {
cout << "True";
}
else cout << "False";
return 0;
}
// Java program to find if there is a subtree with
// given sum
import java.util.HashMap;
import java.util.Stack;
class Node {
int val;
Node left, right;
Node(int x) {
val = x;
left = null;
right = null;
}
}
// Function to check if there is a subtree
// with the given sum
class GfG {
static boolean hasSubtreeWithGivenSum(Node root, int target) {
if (root == null)
return false;
// Map to store the prefix sums
HashMap<Integer, Integer> prefixSumMap = new HashMap<>();
prefixSumMap.put(0, 1);
int sum = 0;
Stack<Node> st = new Stack<>();
st.push(root);
// Iterative depth-first traversal
while (!st.isEmpty()) {
Node curr = st.pop();
// Update the running sum with the current node's value
sum += curr.val;
// Check if we have encountered (sum - target) before
if (prefixSumMap.containsKey(sum - target)) {
return true;
}
// Add the current running sum to the map
prefixSumMap.put(sum, 1);
// Traverse the right and left children
if (curr.right != null) {
st.push(curr.right);
}
if (curr.left != null) {
st.push(curr.left);
}
}
// If no subtree with the given sum is found
return false;
}
public static void main(String[] args) {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
Node root = new Node(1);
root.left = new Node(3);
root.right = new Node(6);
root.left.left = new Node(5);
root.left.right = new Node(9);
root.right.left = new Node(8);
System.out.println(hasSubtreeWithGivenSum(root, 17));
}
}
# Python program to find if there is a subtree with
# given sum
class Node:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
# Function to check if there is a subtree with the given sum
def has_subtree_with_given_sum(root, target):
if root is None:
return False
# Map to store the prefix sums
prefix_sum_map = {0: 1}
sum = 0
stack = [root]
# Iterative depth-first traversal
while stack:
curr = stack.pop()
# Update the running sum with the current node's value
sum += curr.val
# Check if we have encountered (sum - target) before
if (sum - target) in prefix_sum_map:
return True
# Add the current running sum to the map
prefix_sum_map[sum] = 1
# Traverse the right and left children
if curr.right:
stack.append(curr.right)
if curr.left:
stack.append(curr.left)
# If no subtree with the given sum is found
return False
# Creating the given binary tree
# 1
# / \
# 3 6
# / \ /
# 5 9 8
root = Node(1)
root.left = Node(3)
root.right = Node(6)
root.left.left = Node(5)
root.left.right = Node(9)
root.right.left = Node(8)
print(has_subtree_with_given_sum(root, 17))
// C# program to find if there is a subtree with
// given sum
using System;
using System.Collections.Generic;
class Node {
public int Val;
public Node Left;
public Node Right;
public Node(int x) {
Val = x;
Left = null;
Right = null;
}
}
// Function to check if there is a subtree
// with the given sum
class GfG {
static bool HasSubtreeWithGivenSum(Node root, int target) {
if (root == null)
return false;
// Map to store the prefix sums
Dictionary<int, int> prefixSumMap = new Dictionary<int, int>();
prefixSumMap[0] = 1;
int sum = 0;
Stack<Node> stack = new Stack<Node>();
stack.Push(root);
// Iterative depth-first traversal
while (stack.Count > 0) {
Node curr = stack.Pop();
// Update the running sum with the current node's value
sum += curr.Val;
// Check if we have encountered (sum - target) before
if (prefixSumMap.ContainsKey(sum - target)) {
return true;
}
// Add the current running sum to the map
prefixSumMap[sum] = 1;
// Traverse the right and left children
if (curr.Right != null) {
stack.Push(curr.Right);
}
if (curr.Left != null) {
stack.Push(curr.Left);
}
}
// If no subtree with the given sum is found
return false;
}
static void Main() {
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
Node root = new Node(1);
root.Left = new Node(3);
root.Right = new Node(6);
root.Left.Left = new Node(5);
root.Left.Right = new Node(9);
root.Right.Left = new Node(8);
Console.WriteLine(HasSubtreeWithGivenSum(root, 17));
}
}
// JavaScript program to find if there is a subtree with
// given sum
class Node {
constructor(x) {
this.val = x;
this.left = null;
this.right = null;
}
}
// Function to check if there is a subtree with the given sum
function hasSubtreeWithGivenSum(root, target) {
if (root === null) return false;
// Map to store the prefix sums
const prefixSumMap = new Map();
prefixSumMap.set(0, 1);
let sum = 0;
const stack = [root];
// Iterative depth-first traversal
while (stack.length > 0) {
const curr = stack.pop();
// Update the running sum with the current node's value
sum += curr.val;
// Check if we have encountered (sum - target) before
if (prefixSumMap.has(sum - target)) {
return true;
}
// Add the current running sum to the map
prefixSumMap.set(sum, 1);
// Traverse the right and left children
if (curr.right) {
stack.push(curr.right);
}
if (curr.left) {
stack.push(curr.left);
}
}
// If no subtree with the given sum is found
return false;
}
// Creating the given binary tree
// 1
// / \
// 3 6
// / \ /
// 5 9 8
const root = new Node(1);
root.left = new Node(3);
root.right = new Node(6);
root.left.left = new Node(5);
root.left.right = new Node(9);
root.right.left = new Node(8);
console.log(hasSubtreeWithGivenSum(root, 17));
Time complexity : O(n), As we are visiting every node once.
Auxiliary Space : O(n)
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