Leetcode-113. Path Sum II
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解题思路:类似于Path Sum 的思路,只有当前节点是叶子节点并且结点的值等于暂时sum的值时才确定这是一条满足要求的路径,并把路径上的节点存入元素为向量的向量中。这要求在递归遍历节点的同时把节点放入一个临时向量中。实现起来的代码需要非常精致的技巧,每个节点的左右节点遍历完毕,需要将其弹出临时向量。
class Solution {public: vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> vvi; vector<int> vi; if(!root) return vvi; getpath(root,sum,vi,vvi); return vvi; } void getpath(TreeNode* root,int sum,vector<int> &tmp,vector<vector<int>> &result){ if(!root->left && !root->right && sum==root->val) { tmp.push_back(root->val); result.push_back(tmp); tmp.pop_back(); //return; } if(root->left){ tmp.push_back(root->val); getpath(root->left,sum-root->val,tmp,result); } if(root->right){ tmp.push_back(root->val); getpath(root->right,sum-root->val,tmp,result); } if(!tmp.empty()) tmp.pop_back(); } };
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