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112_path_sum.cpp
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112_path_sum.cpp
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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.
/**
* Definition for a binary tree node.
* 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 {
public:
bool hasPathSum(TreeNode* root, int targetSum) {
if (!root)
return false;
if (!root->left && !root->right)
return targetSum - root->val == 0;
return hasPathSum(root->left, targetSum - root->val) || hasPathSum(root->right, targetSum - root->val);
}
};