原题链接：添加链接描述



DFS

/**
 * 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 r=false;
    void dfs(TreeNode* root, int sum, int targetSum)
    {
        if(root->left==nullptr && root->right==nullptr)
        {
            if(sum==targetSum) r=true;
            return;
        }
        if(root->left) dfs(root->left,sum+root->left->val,targetSum);
        if(root->right) dfs(root->right,sum+root->right->val,targetSum);
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root==nullptr) return false;
        dfs(root,root->val,targetSum);
        return r;
    }
};

递归

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