For 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.    

/**
 * 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)
        {
            return false;
        }
        if (root->left == NULL && root->right == NULL && root->val == sum)
        {
            return true;
        }
        else
        {
            return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
        }
    }
};

4. 反思

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