LeetCode: Path Sum II
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思路:和前面一题一样,不过这次需要记录具体路径,在前面中,是通过空节点来进行递归终止的,这样会形成一个解路径重复记录,如一个叶子节点,如果满足条件,则会记录两次,因为左右节点都为空,搜索左节点会记录一次解,搜索右节点也会记录一次,所以,这里用叶子节点进行递归终止条件,其他思路和前面一题一样。
code:
class Solution {public: bool dfs(TreeNode *root,int curSum,int sum,vector<int> curPath, vector<vector<int> > &ret){ if(root->left == NULL && root->right == NULL){ if(curSum + root->val == sum){ curPath.push_back(root->val); ret.push_back(curPath); return true; } return false; } vector<int> t1(curPath); t1.push_back(root->val); bool left = false, right = false; if(root->left != NULL) left = dfs(root->left,curSum+root->val,sum,t1,ret); if(root->right != NULL) right = dfs(root->right,curSum+root->val,sum,t1,ret); return left || right; } vector<vector<int> > pathSum(TreeNode *root, int sum) { vector<vector<int> > ret; vector<int> curRet; if(root == NULL) return ret; dfs(root,0,sum,curRet,ret); return ret; }};
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