pathSumDFS.cpp

112. Path Sum
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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.
  
  
  class Solution {
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(!root) return false; //root doesn' exit then false
        
        if(!root ->left && !root->right) // both left and right have value 
            return root->val ==targetSum; // achieved target then return
     return hasPathSum(root ->left , targetSum - root ->val) || hasPathSum(root ->right ,targetSum - root->val );
    }
};