C++ 解法, 执行用时: 3ms, 内存消耗: 768KB, 提交时间: 2022-07-26

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool dfs(TreeNode* node, int num, int target) {
        int n = num + node->val;
        bool l = false, r = false;
        if (node->left != nullptr) {
            l = true;
        } 
        if (node->right != nullptr) {
            r = true;
        }
        if (!l && !r) {
            return n == target;
        }
        
        bool ret = false;
        if (l) {
            ret = ret | dfs(node->left, n, target);
        }
        if (r) {
            ret = ret | dfs(node->right, n, target);
        }
        return ret;
    }
    bool hasPathSum(TreeNode* root, int sum) {
        // write code here
        if (root == nullptr) {
            return false;
        }
        return dfs(root, 0, sum);
            
        
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 768KB, 提交时间: 2022-06-19

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool hasPathSum(TreeNode* root, int sum) {
        // write code here
        if(root == NULL)
        {
            return false;
        }
        if(root->val == sum && root->left == NULL && root->right == NULL)
        {
            return true;
        }
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 768KB, 提交时间: 2022-06-16

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    int dfs(TreeNode* node, int pre,int res)
    {
        if(node==nullptr) return 0;
        int sum=pre+node->val;
        if(node->left==nullptr&&node->right==nullptr) 
        {
            if(sum==res) return 1;
            else return 0;
        }
            
        return dfs(node->left,sum,res)|dfs(node->right,sum,res);
        
    }
    bool hasPathSum(TreeNode* root, int sum) {
        if(root==nullptr) return false;
       return dfs(root,0,sum);
       
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 768KB, 提交时间: 2022-03-11

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool hasPathSum(TreeNode* root, int sum) {
        // write code here
        if(!root) return false;
        if(!root->left && !root->right){
            if(sum==root->val) return true;
            else return false;
        } 
        bool ans = hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
        return ans;
    }

};

C++ 解法, 执行用时: 3ms, 内存消耗: 768KB, 提交时间: 2021-11-29

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    bool preOrder(TreeNode *node, int sum, int ans)
    {
        if(!node)
            return false;
        ans += node->val;
        if(!node->left && !node->right && sum == ans)
            return true;
        return preOrder(node->left, sum, ans)||
            preOrder(node->right, sum ,ans);
    }
    bool hasPathSum(TreeNode* root, int sum) {
        // write code here
        if(!root)
            return false;
        return preOrder(root, sum, 0);
    }
};