【Leetcode】Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:Given the below binary tree and
sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
思路:与【Leetcode】Path
Sum 不同的是。此题须要保存全部的路径,因此,搜索完左子树的路径后应该继续搜索右子树的路径。
/** * Definition for binary tree * 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>> result; vector<int> path; pathSumHelper(result, path, root, sum); return result; } private: void pathSumHelper(vector<vector<int>> &result, vector<int> &path, TreeNode *node, int sum) { if(node == NULL) return; if(node->left == NULL && node->right == NULL && node->val == sum) { path.push_back(node->val); result.push_back(path); path.pop_back(); return; } path.push_back(node->val); pathSumHelper(result, path, node->left, sum - node->val); pathSumHelper(result, path, node->right, sum - node->val); path.pop_back(); } };
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