剑指offer JZ34 二叉树中和为某一值的路径(二)

原题链接：剑指offer JZ34 二叉树中和为某一值的路径(二)

开始自作主张在dfs里写了个剪枝：if(sum>=expectNumber) return ;，，，，题目里说了节点值有可能为负数，把剪枝去掉就对了

class Solution {
public:
    void dfs(vector>& res,vector& tmp,TreeNode* root,int sum,int expectNumber)
    {
        if(!root->left && !root->right)
        {
            if(sum== expectNumber)
                res.push_back(tmp);
            return ;
        }
        if(root->left)
        {
            tmp.push_back((root->left)->val);
            dfs(res,tmp,root->left,sum+(root->left)->val,expectNumber);
            tmp.pop_back();
        }
        if(root->right)
        {
            tmp.push_back((root->right)->val);
            dfs(res,tmp,root->right,sum+(root->right)->val,expectNumber);
            tmp.pop_back();
        }
        
    }
    vector> FindPath(TreeNode* root,int expectNumber) {
        vector> res;
        vector tmp;
        if(!root)
            return res;
        tmp.push_back(root->val);
        dfs(res,tmp,root,root->val,expectNumber);
        return res;
    }
};