题解 | #二叉树中和为某一值的路径(三)#

比较低效率的方法：层序遍历+dfs

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
private:
    int res = 0;
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 sum int整型 
     * @return int整型
     */
    int FindPath(TreeNode* root, int sum) {
        // write code here
        if(root == nullptr){
            return 0;
        }
        
        queue<TreeNode*> Q;
        Q.push(root);
        
        while(!Q.empty()){
            int cur_size = Q.size();
            for(int i=0; i<cur_size; i++){
                auto node = Q.front();
                Q.pop();
                dfs(node, sum, 0);
                if(node->left) Q.push(node->left);
                if(node->right) Q.push(node->right);
            }
        }
        return res;
    }
    
    void dfs(TreeNode* root, int sum, int cur_sum){
        if(root == nullptr){
            return;
        }
        cur_sum += root->val;
        
        if(sum == cur_sum){
            res += 1;
        }

        dfs(root->left, sum, cur_sum);
        dfs(root->right, sum, cur_sum);
        
    }
    
};