题解 | #牛的奶量统计II#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param targetSum int整型 
     * @return bool布尔型
     */
    
    bool dfs(TreeNode* root, int targetSum)
    {
        if(!root)
        {
            return false;
        }

        targetSum -= root->val;

        // 到达子节点
        if(!root->left && !root->right)
        {
            // cout << root->val << ", " << targetSum-root->val << endl;
            if(targetSum == 0)
                return true;
            else
                return false;
        }

        return dfs(root->left, targetSum) || dfs(root->right, targetSum);
    }

    bool hasPathSumII(TreeNode* root, int targetSum) {
        // write code here
        // 注意题目要求是任意节点到其子节点，则起始节点可以不是根节点;
        if(!root)
        {
            return false;
        }

        if(dfs(root, targetSum))
            return true;

        return hasPathSumII(root->left, targetSum) || hasPathSumII(root->right, targetSum);
    }
};