题解 | #二叉树中和为某一值的路径(三)#

//思路具体看题解。我是参考题解思路写的。
class Solution {
public:

    void solve(TreeNode* root,int sum,int &count)
    {
        if(root==nullptr)
            return;
        if(root->val==sum)
        {
            count++;
        }
        solve(root->left, sum-root->val, count);
        solve(root->right, sum-root->val, count);
    }

    int FrontOrder(TreeNode* root,int sum)
    {
        if(root==nullptr)
            return 0;
        int current=0;
        solve(root, sum, current);//处理当前节点
        int left=0;
        int right=0;
        if(root->left)
        {
            left=FrontOrder(root->left, sum);//递归两个孩子
        }
        if(root->right)
        {
            right=FrontOrder(root->right, sum);
        }
        return left+right+current;
    }

    int FindPath(TreeNode* root, int sum) {
        return FrontOrder(root, sum);
    }
};