牛客OJ：二叉树和为某值的路径

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/ShellDawn/article/details/89048315

BFS：

struct Node{
    TreeNode* p;
    int sum;
    vector<int> v;
};

class Solution {
public:
    vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
        if(root == NULL) return *(new vector<vector<int> >);
        vector<vector<int> > res;
        queue<Node> q;
        Node t;
        t.p = root;
        t.sum = root->val;
        t.v.push_back(root->val);
        q.push(t);
        while(!q.empty()){
            t = q.front();
            q.pop();
            if( t.p->left == NULL && t.p->right == NULL){
                if(t.sum == expectNumber){
                    res.push_back(t.v);
                }
                continue;
            }
            if(t.p->left != NULL){
                Node t1;
                t1.p = t.p->left;
                t1.sum = t.sum + t1.p->val;
                t1.v = t.v;
                t1.v.push_back(t1.p->val);
                q.push(t1);
            }
            if(t.p->right != NULL){
                Node t2;
                t2.p = t.p->right;
                t2.sum = t.sum + t2.p->val;
                t2.v = t.v;
                t2.v.push_back(t2.p->val);
                q.push(t2);
            }
        }
        vector<vector<int> > ans;
        int l = res.size()-1;
        for(int i=l;i>=0;i--){
            ans.push_back(res[i]);
        }
        return ans;
    }
};