Q12_path_sum.cpp

// Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

// A leaf is a node with no children.

// Example 1:
// Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
// Output: true
// Explanation: The root-to-leaf path with the target sum is shown.

// Example 2:
// Input: root = [1,2,3], targetSum = 5
// Output: false
// Explanation: There two root-to-leaf paths in the tree:
// (1 --> 2): The sum is 3.
// (1 --> 3): The sum is 4.
// There is no root-to-leaf path with sum = 5.

// Example 3:
// Input: root = [], targetSum = 0
// Output: false
// Explanation: Since the tree is empty, there are no root-to-leaf paths.

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

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
private:
    bool findPath(TreeNode* root, int leftSum) {
        if(!root) return false;
        if(!root->left && !root->right) return leftSum - root->val == 0;

        bool left = findPath(root->left, leftSum - root->val);
        bool right = findPath(root->right, leftSum - root->val);

        return (left || right);
    }
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        return findPath(root, targetSum);
    }
};

// Time Complexity  : O(n) 
// Space Complexity : O(h)
// where h is the height of the tree