剑指offer JZ34 二叉树和为某一值的路径(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

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    void dfs(vector<vector<int>>& res,vector<int>& 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<vector<int>> FindPath(TreeNode* root,int expectNumber) {
        vector<vector<int>> res;
        vector<int> tmp;
        if(!root)
            return res;
        tmp.push_back(root->val);
        dfs(res,tmp,root,root->val,expectNumber);
        return res;
    }
};

这篇好文章是转载于：学新通技术网