Subtree with given sum in a Binary Tree

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 :

// For above tree
Input : sum = 17
Output: “Yes”
// sum of all nodes of subtree {3, 5, 9} = 17
Input : sum = 11
Output: “No”
// no subtree with given sum exist

The idea is to traverse the tree in a Postorder fashion because here we have to think bottom-up. First, calculate the sum of the left subtree then the right subtree, and check if sum_left + sum_right + cur_node = sum is satisfying the condition that means any subtree with a given sum exists. Below is the recursive implementation of the algorithm. 

C++

// C++ program to find if there is a subtree with
// given sum
#include<bits/stdc++.h>
using namespace std;

/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct Node
{
    int data;
    struct Node* left, *right;
};

/* utility that allocates a new node with the
given data and NULL left and right pointers. */
struct Node* newnode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right  = NULL;
    return (node);
}

// function to check if there exist any subtree with given sum
// cur_sum  --> sum of current subtree from ptr as root
// sum_left --> sum of left subtree from ptr as root
// sum_right --> sum of right subtree from ptr as root
bool sumSubtreeUtil(struct Node *ptr, int *cur_sum, int sum)
{
    // base condition
    if (ptr == NULL)
    {
        *cur_sum = 0;
        return false;
    }

    // Here first we go to left sub-tree, then right subtree
    // then first we calculate sum of all nodes of subtree
    // having ptr as root and assign it as cur_sum
    // cur_sum = sum_left + sum_right + ptr->data
    // after that we check if cur_sum == sum
    int sum_left = 0, sum_right = 0;
    return ( sumSubtreeUtil(ptr->left, &sum_left, sum) ||
             sumSubtreeUtil(ptr->right, &sum_right, sum) ||
        ((*cur_sum = sum_left + sum_right + ptr->data) == sum));
}

// Wrapper over sumSubtreeUtil()
bool sumSubtree(struct Node *root, int sum)
{
    // Initialize sum of subtree with root
    int cur_sum = 0;

    return sumSubtreeUtil(root, &cur_sum, sum);
}

// driver program to run the case
int main()
{
    struct Node *root = newnode(8);
    root->left    = newnode(5);
    root->right   = newnode(4);
    root->left->left = newnode(9);
    root->left->right = newnode(7);
    root->left->right->left = newnode(1);
    root->left->right->right = newnode(12);
    root->left->right->right->right = newnode(2);
    root->right->right = newnode(11);
    root->right->right->left = newnode(3);
    int sum = 22;

    if (sumSubtree(root, sum))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find if there
// is a subtree with given sum
import java.util.*;
class GFG
{

/* A binary tree node has data,
pointer to left child and a
pointer to right child */
static class Node
{
    int data;
    Node left, right;
}

static class INT
{
    int v;
    INT(int a)
    {
        v = a;
    }
}

/* utility 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);
}

// function to check if there exist
// any subtree with given sum
// cur_sum -. sum of current subtree
//            from ptr as root
// sum_left -. sum of left subtree
//             from ptr as root
// sum_right -. sum of right subtree
//              from ptr as root
static boolean sumSubtreeUtil(Node ptr,
                              INT cur_sum,
                              int sum)
{
    // base condition
    if (ptr == null)
    {
        cur_sum = new INT(0);
        return false;
    }

    // Here first we go to left
    // sub-tree, then right subtree
    // then first we calculate sum
    // of all nodes of subtree having
    // ptr as root and assign it as
    // cur_sum. (cur_sum = sum_left +
    // sum_right + ptr.data) after that
    // we check if cur_sum == sum
    INT sum_left = new INT(0),
        sum_right = new INT(0);
    return (sumSubtreeUtil(ptr.left, sum_left, sum) ||
            sumSubtreeUtil(ptr.right, sum_right, sum) ||
        ((cur_sum.v = sum_left.v +
                      sum_right.v + ptr.data) == sum));
}

// Wrapper over sumSubtreeUtil()
static boolean sumSubtree(Node root, int sum)
{
    // Initialize sum of
    // subtree with root
    INT cur_sum = new INT( 0);

    return sumSubtreeUtil(root, cur_sum, sum);
}

// Driver Code
public static void main(String args[])
{
    Node root = newnode(8);
    root.left = newnode(5);
    root.right = newnode(4);
    root.left.left = newnode(9);
    root.left.right = newnode(7);
    root.left.right.left = newnode(1);
    root.left.right.right = newnode(12);
    root.left.right.right.right = newnode(2);
    root.right.right = newnode(11);
    root.right.right.left = newnode(3);
    int sum = 22;

    if (sumSubtree(root, sum))
        System.out.println( "Yes");
    else
        System.out.println( "No");
}
}

// This code is contributed
// by Arnab Kundu

Python3

# Python3 program to find if there is a
# subtree with given sum

# Binary Tree Node
""" utility that allocates a newNode
with the given key """
class newnode:

    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None

# function to check if there exist any
# subtree with given sum
# cur_sum -. sum of current subtree
#            from ptr as root
# sum_left -. sum of left subtree from
#             ptr as root
# sum_right -. sum of right subtree
#              from ptr as root
def sumSubtreeUtil(ptr,cur_sum,sum):

    # base condition
    if (ptr == None):
        cur_sum[0] = 0
        return False

    # Here first we go to left sub-tree,
    # then right subtree then first we
    # calculate sum of all nodes of subtree
    # having ptr as root and assign it as cur_sum
    # cur_sum = sum_left + sum_right + ptr.data
    # after that we check if cur_sum == sum
    sum_left, sum_right = [0], [0]
    x=sumSubtreeUtil(ptr.left, sum_left, sum)
    y=sumSubtreeUtil(ptr.right, sum_right, sum)
    cur_sum[0] = (sum_left[0] +
                  sum_right[0] + ptr.data)
    return ((x or y)or (cur_sum[0] == sum))

# Wrapper over sumSubtreeUtil()
def sumSubtree(root, sum):

    # Initialize sum of subtree with root
    cur_sum = [0]

    return sumSubtreeUtil(root, cur_sum, sum)

# Driver Code
if __name__ == '__main__':

    root = newnode(8)
    root.left = newnode(5)
    root.right = newnode(4)
    root.left.left = newnode(9)
    root.left.right = newnode(7)
    root.left.right.left = newnode(1)
    root.left.right.right = newnode(12)
    root.left.right.right.right = newnode(2)
    root.right.right = newnode(11)
    root.right.right.left = newnode(3)
    sum = 22

    if (sumSubtree(root, sum)) :
        print("Yes" )
    else:
        print("No")

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

using System;

// C# program to find if there
// is a subtree with given sum
public class GFG
{

/* A binary tree node has data,
pointer to left child and a
pointer to right child */
public class Node
{
    public int data;
    public Node left, right;
}

public class INT
{
    public int v;
    public INT(int a)
    {
        v = a;
    }
}

/* utility that allocates a new
node with the given data and
null left and right pointers. */
public static Node newnode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}

// function to check if there exist
// any subtree with given sum
// cur_sum -. sum of current subtree
//         from ptr as root
// sum_left -. sum of left subtree
//             from ptr as root
// sum_right -. sum of right subtree
//             from ptr as root
public static bool sumSubtreeUtil(Node ptr, INT cur_sum, int sum)
{
    // base condition
    if (ptr == null)
    {
        cur_sum = new INT(0);
        return false;
    }

    // Here first we go to left
    // sub-tree, then right subtree
    // then first we calculate sum
    // of all nodes of subtree having
    // ptr as root and assign it as
    // cur_sum. (cur_sum = sum_left +
    // sum_right + ptr.data) after that
    // we check if cur_sum == sum
    INT sum_left = new INT(0), sum_right = new INT(0);
    return (sumSubtreeUtil(ptr.left, sum_left, sum)
            || sumSubtreeUtil(ptr.right, sum_right, sum)
            || ((cur_sum.v = sum_left.v + sum_right.v + ptr.data) == sum));
}

// Wrapper over sumSubtreeUtil()
public static bool sumSubtree(Node root, int sum)
{
    // Initialize sum of
    // subtree with root
    INT cur_sum = new INT(0);

    return sumSubtreeUtil(root, cur_sum, sum);
}

// Driver Code
public static void Main(string[] args)
{
    Node root = newnode(8);
    root.left = newnode(5);
    root.right = newnode(4);
    root.left.left = newnode(9);
    root.left.right = newnode(7);
    root.left.right.left = newnode(1);
    root.left.right.right = newnode(12);
    root.left.right.right.right = newnode(2);
    root.right.right = newnode(11);
    root.right.right.left = newnode(3);
    int sum = 22;

    if (sumSubtree(root, sum))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by Shrikant13

Javascript

<script>
// javascript program to find if there
// is a subtree with given sum

    /*
     * A binary tree node has data, pointer to left child and a pointer to right
     * child
     */
     class Node {
        constructor(){
        this.data = 0;
        this.left = null;
        this.right = null;
        }
    }

     class INT {

        constructor(a) {
            this.v = a;
        }
    }

    /*
     * utility that allocates a new node with the given data and null left and right
     * pointers.
     */
    function newnode(data) {
        var node = new Node();
        node.data = data;
        node.left = node.right = null;
        return (node);
    }

    // function to check if there exist
    // any subtree with given sum
    // cur_sum -. sum of current subtree
    // from ptr as root
    // sum_left -. sum of left subtree
    // from ptr as root
    // sum_right -. sum of right subtree
    // from ptr as root
    function sumSubtreeUtil( ptr,  cur_sum , sum)
    {

        // base condition
        if (ptr == null)
        {
            cur_sum = new INT(0);
            return false;
        }

        // Here first we go to left
        // sub-tree, then right subtree
        // then first we calculate sum
        // of all nodes of subtree having
        // ptr as root and assign it as
        // cur_sum. (cur_sum = sum_left +
        // sum_right + ptr.data) after that
        // we check if cur_sum == sum
        var sum_left = new INT(0), sum_right = new INT(0);
        return (sumSubtreeUtil(ptr.left, sum_left, sum) || sumSubtreeUtil(ptr.right, sum_right, sum)
                || ((cur_sum.v = sum_left.v + sum_right.v + ptr.data) == sum));
    }

    // Wrapper over sumSubtreeUtil()
    function sumSubtree( root, sum)
    {

        // Initialize sum of
        // subtree with root
        var cur_sum = new INT(0);

        return sumSubtreeUtil(root, cur_sum, sum);
    }

    // Driver Code

        var root = newnode(8);
        root.left = newnode(5);
        root.right = newnode(4);
        root.left.left = newnode(9);
        root.left.right = newnode(7);
        root.left.right.left = newnode(1);
        root.left.right.right = newnode(12);
        root.left.right.right.right = newnode(2);
        root.right.right = newnode(11);
        root.right.right.left = newnode(3);
        var sum = 22;

        if (sumSubtree(root, sum))
            document.write("Yes");
        else
            document.write("No");

// This code is contributed by Rajput-Ji
</script>

Output

Yes

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.

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Approach 2:- 

•  Initialize a hash map mp to store the frequency of each sum encountered along the root-to-leaf paths of the binary tree.
• Add a dummy entry in the map with sum 0 to handle the case where the root itself has the target sum.
•  Initialize a stack st and push the root node onto it.
• While the stack is not empty, pop the top node curr from the stack.
  • Update the sum as sum + curr->val.
  • Check if the difference sum – targetSum exists in the hash map. If it does, then return true.
  • Otherwise, add the current sum sum to the hash map with a frequency of 1.
  •  If the current node has a right child, push it onto the stack.
  • If the current node has a left child, push it onto the stack.
  • If the stack becomes empty, return false.

C++

#include <bits/stdc++.h>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x)
        : val(x)
        , left(NULL)
        , right(NULL)
    {
    }
};

bool hasTargetSum(TreeNode* root, int targetSum)
{
    unordered_map<int, int> mp;
    mp[0] = 1; // adding a dummy sum to handle the case
               // where root itself has the target sum
    int sum = 0;
    stack<TreeNode*> st;
    st.push(root);
    while (!st.empty()) {
        TreeNode* curr = st.top();
        st.pop();
        sum += curr->val;
        if (mp.find(sum - targetSum) != mp.end()) {
            return true;
        }
        mp[sum] = 1;
        if (curr->right) {
            st.push(curr->right);
        }
        if (curr->left) {
            st.push(curr->left);
        }
    }
    return false;
}

int main()
{
    /*
           5
         /   \
        4     8
       /     / \
      11    13  4
     /  \       \
    7    2       1
    */
    TreeNode* root = new TreeNode(5);
    root->left = new TreeNode(4);
    root->left->left = new TreeNode(11);
    root->left->left->left = new TreeNode(7);
    root->left->left->right = new TreeNode(2);
    root->right = new TreeNode(8);
    root->right->left = new TreeNode(13);
    root->right->right = new TreeNode(4);
    root->right->right->right = new TreeNode(1);
    int targetSum = 22;
    if (hasTargetSum(root, targetSum)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;

// Create tree node
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

class Solution {
    // Function to check the target
    public boolean hasTargetSum(TreeNode root,
                                int targetSum)
    {
        HashMap<Integer, Integer> map = new HashMap<>();
        map.put(0,
                1); // adding a dummy sum to handle the case
                    // where root itself has the target sum
        int sum = 0;
        // Using stack to push node
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);

        // Looping the stack
        while (!stack.empty()) {
            TreeNode curr = stack.pop();
            sum += curr.val;
            if (map.containsKey(
                    sum - targetSum)) { // checking target
                return true;
            }
            map.put(sum, 1);
            if (curr.right
                != null) { // calling the right node
                stack.push(curr.right);
            }
            if (curr.left
                != null) { // calling the left node
                stack.push(curr.left);
            }
        }
        return false;
    }

    public static void main(String[] args)
    {

        // Taken Tree
        /*
                 5
               /   \
              4     8
             /     / \
            11    13  4
           /  \       \
          7    2       1
          */
        // Input tree
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(4);
        root.left.left = new TreeNode(11);
        root.left.left.left = new TreeNode(7);
        root.left.left.right = new TreeNode(2);
        root.right = new TreeNode(8);
        root.right.left = new TreeNode(13);
        root.right.right = new TreeNode(4);
        root.right.right.right = new TreeNode(1);
        int targetSum = 22;

        Solution solution = new Solution();
        if (solution.hasTargetSum(root, targetSum)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

Python3

# Python implementation of above approach
from collections import defaultdict

# Create tree node
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

  # Function to check the target
def hasTargetSum(root, targetSum):
    mp = defaultdict(int)
    mp[0] = 1  # adding a dummy sum to handle the case
               # where root itself has the target sum

    stack = [root]   # Using stack to push node
    sum = 0

     # Looping the stack
    while stack:
        curr = stack.pop()
        sum += curr.val
        if sum - targetSum in mp:    # checking target
            return True
        mp[sum] = 1
        if curr.right:
            stack.append(curr.right)  # calling the right node
        if curr.left:
            stack.append(curr.left)   # calling the left node
    return False

"""
       5
     /   \
    4     8
   /     / \
  11    13  4
 /  \       \
7    2       1
"""

# Driver code

root = TreeNode(5)
root.left = TreeNode(4)
root.left.left = TreeNode(11)
root.left.left.left = TreeNode(7)
root.left.left.right = TreeNode(2)
root.right = TreeNode(8)
root.right.left = TreeNode(13)
root.right.right = TreeNode(4)
root.right.right.right = TreeNode(1)
targetSum = 22
if hasTargetSum(root, targetSum):
    print("Yes")
else:
    print("No")

C#

using System;
using System.Collections.Generic;

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) { val = x; }
}

public class Solution {
    public bool HasTargetSum(TreeNode root, int targetSum)
    {
        Dictionary<int, int> mp
            = new Dictionary<int, int>();
        mp[0] = 1; // adding a dummy sum to handle the case
                   // where root itself has the target sum
        int sum = 0;
        Stack<TreeNode> st = new Stack<TreeNode>();
        st.Push(root);
        while (st.Count != 0) {
            TreeNode curr = st.Pop();
            sum += curr.val;
            if (mp.ContainsKey(
                    sum - targetSum)) { // check if the
                                        // difference exists
                                        // in the dictionary
                return true;
            }
            mp[sum] = 1; // add the sum to the dictionary
            if (curr.right != null) {
                st.Push(
                    curr.right); // add the right child to
                                 // the stack if it exists
            }
            if (curr.left != null) {
                st.Push(
                    curr.left); // add the left child to the
                                // stack if it exists
            }
        }
        return false;
    }
}

public class Program {
    static void Main(string[] args)
    {
        /*
               5
             /   \
            4     8
           /     / \
          11    13  4
         /  \       \
        7    2       1
        */
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(4);
        root.left.left = new TreeNode(11);
        root.left.left.left = new TreeNode(7);
        root.left.left.right = new TreeNode(2);
        root.right = new TreeNode(8);
        root.right.left = new TreeNode(13);
        root.right.right = new TreeNode(4);
        root.right.right.right = new TreeNode(1);
        int targetSum = 22;
        Solution sol = new Solution();
        if (sol.HasTargetSum(
                root,
                targetSum)) { // check if the target sum
                              // exists in the tree
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
}

Javascript

// Definition for a binary tree node.
function TreeNode(val) {
    this.val = val;
    this.left = this.right = null;
}

function hasTargetSum(root, targetSum) {
    const map = new Map();
    map.set(0, 1); // adding a dummy sum to handle the case
    // where root itself has the target sum
    let sum = 0;
    const stack = [];
    stack.push(root);
    while (stack.length > 0) {
        const curr = stack.pop();
        sum += curr.val;
        if (map.has(sum - targetSum)) {
            return true;
        }
        map.set(sum, 1);
        if (curr.right) {
            stack.push(curr.right);
        }
        if (curr.left) {
            stack.push(curr.left);
        }
    }
    return false;
}
/*
5
/
4 8
/ /
11 13 4
/ \
7 2 1
*/
const root = new TreeNode(5);
root.left = new TreeNode(4);
root.left.left = new TreeNode(11);
root.left.left.left = new TreeNode(7);
root.left.left.right = new TreeNode(2);
root.right = new TreeNode(8);
root.right.left = new TreeNode(13);
root.right.right = new TreeNode(4);
root.right.right.right = new TreeNode(1);

const targetSum = 22;
if (hasTargetSum(root, targetSum)) {
    console.log("Yes"); // expected output: "Yes"
} else {
    console.log("No"); // expected output: "No"
}
// This code is contributed by sarojmcy2e

Output

Yes

Time complexity : – O(N)
Auxiliary Space :- O(N)