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题目：
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] ]
代码：
/** * 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>> pathSum(TreeNode* root, int sum) { vector<vector<int> > ret; // Terminal conditon 1 : to the null (not leaf node) // even if sum equals zero at the time, becasue not leaf node // so it is also not the available solution path if (!root) return ret; // Terminal condition 2 : leaf node if ( !root->left && !root->right ) { // if leaf node‘s val equal sum if ( sum==root->val ) { vector<int> tmp; tmp.insert(tmp.begin(),root->val); ret.push_back(tmp); } return ret; } // not leaf node : move forward to left and right vector<vector<int> > l = Solution::pathSum(root->left, sum - root->val); vector<vector<int> > r = Solution::pathSum(root->right, sum - root->val); for ( size_t i = 0; i<l.size(); ++i ) { l[i].insert(l[i].begin(), root->val); ret.push_back(l[i]); } for ( size_t i = 0; i<r.size(); ++i ) { r[i].insert(r[i].begin(), root->val); ret.push_back(r[i]); } return ret; } };
tips：
与Path Sum思路类似（http://www.cnblogs.com/xbf9xbf/p/4508964.html）
不同的地方是：只要不是leaf node，left和right两边的情况都要考虑。
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并且有个地方需要缕清思路：如果递归时遇上root==NULL，直接返回空的ret是否合理？如果此时的sum==0呢？
root==NULL有以下三种情况
1. 如果整棵树是空树：即使sum==0也不满足条件
2. 某个非leaf node的left（或right）为空：则即使此时sum==0，往left(或right)方向走会得到root=NULL，因为此时root不是leaf node也不成立
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4510667.html
