标签：
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] ]思路：DFS. 如果root为空，直接返回空数组，否则调用help函数。help函数将传入的root节点值压入cur数组，更新当前数组cur的和值now，如果root为叶子节点，且cur数组和为target，则将cur数组压入ret。如果左子结点不为空，递归调用 help(root->left,cur,now,target)，右子结点不为空，递归调用 help(root->right,cur,now,target)。请注意，cur数组不能传入引用参数，应该传入形参。每个递归函数在运行过程中对cur数组的操作都是各自为政，互不干涉的。
/*** 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>> ret;void help(TreeNode *root,vector<int> cur,int now,int target){now+=root->val;cur.push_back(root->val);if(!root->left&&!root->right){if(now==target)ret.push_back(cur);return;}if(root->left)help(root->left,cur,now,target);if(root->right)help(root->right,cur,now,target);}vector<vector<int>> pathSum(TreeNode* root, int sum) {if(!root)return ret;vector<int> temp;help(root,temp,0,sum);return ret;}};
标签：
原文地址：http://www.cnblogs.com/zhoudayang/p/5043195.html
