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 Sum(TreeNode* root,int num,int target)
    {
        // 叶节点，不能使用叶节点之后的节点进行判断
        // 1 2 这种情况
        // 如果向根节点的右树发展就是对的，可是这就与题目相悖
        num+=root->val;
        if(num==target && !root->left && !root->right) return true;
        if((!root->left && !root->right) && num != target) return false;
        if(root->left && Sum(root->left,num,target)) return true;
        if(root->right && Sum(root->right,num,target)) return true;
        return false;
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root==nullptr) return false;
        return Sum(root,0,targetSum);
    }
};

优秀代码

class Solution {
public:
    bool Sum(TreeNode* root,int num,int target)
    {
        if(root==nullptr) return false;
        num+=root->val;
        if(!root->left && !root->right) 
        {
            if(num== target) return true;
            return false;
        }
        return Sum(root->left,num,target) || Sum(root->right,num,target);
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root==nullptr) return false;
        return Sum(root,0,targetSum);
    }
};