1 不仅仅局限有无，针对所有情况的枚举处理

语言的差异 Java vs C++

  • root.val ,not root->val ;
  • null not C++ NULL
  • java vector结构的细微差异，链表添加链表！，链表remove注意传入int的意义
  • {},0的边界输入

2 code

本轮重点是java

有2个版本




public class Solution {

    //全局变量
    public ArrayList<Integer> SmallList = new ArrayList<Integer>() ;
    public ArrayList<ArrayList<Integer>>  BigList = new ArrayList<ArrayList<Integer>> ();

    public ArrayList<ArrayList<Integer>> FindPath(TreeNode root,int expectNumber) {
        //001 check
        if(root == null){
            //deal 边界
            return BigList;
        }
        //002 deep search
        deepSearch(root,expectNumber);

        //容易忘掉
        return BigList;
    }

    public void deepSearch(TreeNode root,int expectNumber){

        //stack push 
        Integer currInteger = Integer.valueOf(root.val);
        SmallList.add(currInteger);

        //found it;
        if(root.left == null && root.right ==null && expectNumber == root.val){
            //BigList.add(SmallList);
            BigList.add(new ArrayList<Integer>(SmallList) );
        }

        //sub routine;
        if(root.left !=null ){
            deepSearch(root.left, expectNumber - root.val);
        }
        if(root.right !=null ){
            deepSearch(root.right, expectNumber - root.val);
        }

        //final hard: stack pop; 容易出错
        SmallList.remove(currInteger);
    }
}


//这个版本 我还没提交过，仅仅参考


class Solution {
public:
    using vvi = vector<vector<int>>;
    using vi = vector<int>;
    void dfs(TreeNode *root, int sum, vi &path, vvi &ret) {
        path.push_back(root->val);
        if (sum == root->val && !root->left && !root->right) {
            ret.push_back(path);
        }
        if (root->left) dfs(root->left, sum - root->val, path, ret);
        if (root->right) dfs(root->right, sum - root->val, path, ret);
        path.pop_back(); // 代码走到这里，表明要回溯，代表当前path中的root节点我已经不需要了
    }
    vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
        vvi ret;
        vi path;
        if (!root) return ret;
        dfs(root, expectNumber, path, ret);
        return ret;
    }
};