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问题:
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:sum = 22
, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5]]解答:
上一题path sum的升级,多加一个记录路径的变量。
但是如果我没做上一题,直接做这一题的话,可能会采取先记录路径,再求和的方法,显然笨拙了很多。
代码:
class Solution {public: vector> pathSum(TreeNode *root, int sum) { vector > re_list; vector result; DFS(root, 0, sum, result, re_list); return re_list; } void DFS(TreeNode *root, int sum, int target, vector result, vector > &re_list) { if(root == NULL) return; if(root->left == NULL && root->right == NULL) { if(target == root->val + sum) { result.push_back(root->val); re_list.push_back(result); } } else { result.push_back(root->val); DFS(root->left, sum+root->val, target, result, re_list); DFS(root->right, sum+root->val, target, result, re_list); } }};
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