题解 | #二叉树中和为某一值的路径(二)#

dfs + 回溯

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    vector<vector<int>> v;
    vector<vector<int>> FindPath(TreeNode* root,int expectNumber) {
        if(!root)return {};
        vector<int> t;
        dfs(root,expectNumber,t,0);
        return v;
    }
    void dfs(TreeNode* r,int e,vector<int> &t,int ans){
        if(!r)return;
        ans += r->val;
        t.push_back(r->val);
        if(ans == e && !r->left && !r->right){
            v.push_back(t);
        }
        dfs(r->left,e,t,ans);
        dfs(r->right,e,t,ans);
        t.pop_back();
    }
};