标签:treenode push oid add dfs out leetcode ret ==
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> res; if(root == NULL) return res; vector<int> add; add.push_back(root->val); DFS(res,add,root,sum,root->val); return res; } void DFS(vector<vector<int>>& res,vector<int>& add,TreeNode*root,int sum,int& he){ if((root->left == NULL)&&(root->right == NULL)){ if(he == sum) res.push_back(add); } else if((root->left != NULL)&&(root->right == NULL)){ add.push_back(root->left->val); he += root->left->val; DFS(res,add,root->left,sum,he); add.pop_back(); he -= root->left->val; } else if((root->left == NULL)&&(root->right != NULL)){ add.push_back(root->right->val); he += root->right->val; DFS(res,add,root->right,sum,he); add.pop_back(); he -= root->right->val; } else if((root->left != NULL)&&(root->right != NULL)){ add.push_back(root->left->val); he += root->left->val; DFS(res,add,root->left,sum,he); add.pop_back(); he -= root->left->val; add.push_back(root->right->val); he += root->right->val; DFS(res,add,root->right,sum,he); add.pop_back(); he -= root->right->val; } return; } };
_虽然代码丑,但比较好理解
class Solution { public: vector<vector<int> > pathSum(TreeNode *root, int sum) { vector<vector<int>> res; vector<int> out; helper(root, sum, out, res); return res; } void helper(TreeNode* node, int sum, vector<int>& out, vector<vector<int>>& res) { if (!node) return; out.push_back(node->val); if (sum == node->val && !node->left && !node->right) { res.push_back(out); } helper(node->left, sum - node->val, out, res); helper(node->right, sum - node->val, out, res); out.pop_back(); } };
——这个和上一题对应
标签:treenode push oid add dfs out leetcode ret ==
原文地址:https://www.cnblogs.com/cunyusup/p/10348257.html