Path Sum（C++路徑總和）

解題思路：

（1）依次判斷左右子樹是否滿足即可，遞歸求解

/**
 * 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 sum) {
        if(root==NULL) 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);
    }
};