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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http://www.cnblogs.com/vincently/p/4130398.html一样的思路，只不过注意保存路径。
时间复杂度O(n),空间复杂度O(lgN)
相关题目：《剑指offer》面试题25
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > pathSum(TreeNode *root, int sum) { 13 vector<vector<int> > result; 14 if (root == NULL) 15 return result; 16 17 vector<int> path; 18 pathSum(root, sum, result, path); 19 20 return result; 21 } 22 23 void pathSum(TreeNode *root, int sum, vector<vector<int> > &result, vector<int> &path) { 24 path.push_back(root->val); 25 26 if (root->left == NULL && root->right == NULL) { 27 if (root->val == sum) { 28 result.push_back(path); 29 } 30 } 31 32 if (root->left) 33 pathSum(root->left, sum - root->val, result, path); 34 35 if (root->right) 36 pathSum(root->right, sum - root->val, result, path); 37 38 path.pop_back(); 39 40 } 41 };
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原文地址：http://www.cnblogs.com/vincently/p/4130504.html
