112. Path Sum

• Difficulty: Easy

• Topics: Tree, Depth-first Search

• Similar Questions:

  • Path Sum II

  • Binary Tree Maximum Path Sum

  • Sum Root to Leaf Numbers

  • Path Sum III

  • Path Sum IV

Problem:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solutions:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == nullptr)    return false;
        if (root->left == nullptr && root->right == nullptr) {
            return root->val == sum;
        }

        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};