关于递归的返回值传递

PUdd2023-11-13 19:56

写leetcode 112.路径总和的时候对返回值的理解加深了

1. 通过成员变量来传递状态

首先是一个通过成员变量来传递状态的代码，没有用到递归返回值传递：

cpp 复制代码

class Solution {
public:
    int temp = 0;
    // 前序遍历 
    void PathSum(TreeNode* cur, int sum, int targetSum){
        sum += cur->val;
        if(cur->left == nullptr && cur->right == nullptr){
            // 说明是叶子结点，sum 到此为止可以判断了。
            if(sum == targetSum){
                temp = 1;
                return;
            }
            else
                return;
        }
        if(cur->left != nullptr) PathSum(cur->left, sum, targetSum);
        if(cur->right != nullptr) PathSum(cur->right, sum, targetSum);
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        int sum = 0;
        if( root == nullptr)
            return 0;
        PathSum(root, sum, targetSum);
        return (temp == 1);
    }
};

2. 误以为"传递"是自动的

然后改成用返回值传递的时候懵了

直接用了PathSum(cur->left, sum, targetSum); 完全没用到它的返回值，以为可以直接return

cpp 复制代码

class Solution {
public:
    // 前序遍历 
    bool PathSum(TreeNode* cur, int sum, int targetSum){
        sum += cur->val;
        if(cur->left == nullptr && cur->right == nullptr && sum == targetSum)
            return 1;
        
        if(cur->left != nullptr) PathSum(cur->left, sum, targetSum);
        if(cur->right != nullptr) PathSum(cur->right, sum, targetSum);
        return 0;
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        int sum = 0;
        if( root == nullptr)
            return 0;
        return PathSum(root, sum, targetSum);
    }
};

3.用递归函数返回布尔值进行传递

小修一下，用到返回值

cpp 复制代码

class Solution {
public:
    // 前序遍历 
    bool PathSum(TreeNode* cur, int sum, int targetSum) {
        sum += cur->val;
        if (cur->left == nullptr && cur->right == nullptr && sum == targetSum) {
            return true;
        }
        
        if (cur->left != nullptr && PathSum(cur->left, sum, targetSum)) return true;
        if (cur->right != nullptr && PathSum(cur->right, sum, targetSum)) return true;

        return false;
    }

    bool hasPathSum(TreeNode* root, int targetSum) {
        if (root == nullptr) {
            return false;
        }
        return PathSum(root, 0, targetSum);
    }
};