Root to Leaf path with target sum Leetcode Solutions  

Difficulty Level Easy
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A binary tree and an integer K are given. Our goal is to return whether there is a root-to-leaf path in the tree such that it’s sum is equal to the target-K. The sum of a path is the sum of all nodes that lie on it.

Root to Leaf path with target sum Leetcode SolutionsPin

    2
    
  /    \
  
1        4
 
       /     \

     1         4

K = 7
Path Present
     2
          
  /     \

1         4

       /     \

     1          3

K = 6
No such path

 

Approach  

The approach is pretty simple. We can check that if at a certain point, the value of a node is what we need, and it is a leaf as well. Then, we can conclude that a path exists from root to this point having target sum. So, it is intuitive that as we recursively jump to any child, we must pass on the remaining sum to complete the target.

This idea works in a single pass of the tree.

Algorithm  

  1. Create a recursive helper function
  2. Helper function has the current node details and the remaining sum
  3. If the current root is NULL,
    • return false, as this node can’t contribute anything.
  4. If the current root is NOT NULL
    • If the sum required is equal to node value and the node is a leaf
      • return true
    • else
      • check if any left or child subtree satisfies the need with required sum: current_sum – node_value
  5. Print the output
See also
Summary Ranges Leetcode Solution

Implementation  

C++ Program to find Root to Leaf path with target sum

#include <bits/stdc++.h>
using namespace std;
struct treeNode
{
    int value;
    treeNode *left , *right;
    treeNode(int x)
    {
        value = x;
        left = NULL;
        right = NULL;
    }
};

bool pathSumEqualToTarget(treeNode* root , int target)
{
    if(!root)
        return false;
    if(target == root->value && !root->left && !root->right)
        return true;
    return pathSumEqualToTarget(root->left , target - root->value) || pathSumEqualToTarget(root->right , target - root->value);
}


int main()
{
    treeNode* root = new treeNode(2);
    root->left = new treeNode(1);
    root->right =  new treeNode(4);
    root->right->left = new treeNode(1);
    root->right->right = new treeNode(4);

    int K = 7;

    if(pathSumEqualToTarget(root , K))
        cout << "Path Present" << '\n';
    else
        cout << "No such path" << '\n';

}

 

Java Program to find Root to Leaf path with target sum

class treeNode
{
    int value;
    treeNode left , right;

    public treeNode(int x)
    {
        value = x;
        left = right = null;
    }
}

class path_sum
{
    public static boolean pathSumEqualToTarget(treeNode root , int target)
    {
        if(root == null)
            return false;
        if(root.value == target && root.left == null && root.right == null)
            return true;
        return pathSumEqualToTarget(root.left , target - root.value) || pathSumEqualToTarget(root.right , target - root.value);
    }


    public static void main(String args[])
    {
        treeNode root = new treeNode(2);
        root.left = new treeNode(1);
        root.right = new treeNode(4);
        root.right.left = new treeNode(1);
        root.right.right = new treeNode(4);

        int K = 7;

        if(pathSumEqualToTarget(root , K))
            System.out.println("Path Present");
        else
            System.out.println("No such path");
    }
}

Complexity Analysis  

Time Complexity to find Root to Leaf path with target sum

O(N), as the program works in a single pass of the binary tree in the worst case.

Space Complexity to find Root to Leaf path with target sum

O(N). Recursion uses auxiliary stack space.

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