# [C++]LeetCode: 91 Path Sum II

2015-01-13

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]
]

Attention:

1. 注意只有存在左右孩子结点时，才递归调用辅助函数，搜索左右子树。

//需要判断是否有左右结点，没有就不递归调用了
if(root->left) pathSum_helper(root->left, newSum, tmp, ret);
if(root->right) pathSum_helper(root->right, newSum, tmp, ret);

2. 当判断当前节点符合迭代终止条件后，首先应先把当前节点&#20540;push进数组，再返回结果。（上一层调用中并没有push当前节点&#20540;）

/如果此时的节点就是叶子节点，并且符合sum条件，先push进tmp，再返回。
if(!root->left && !root->right && root->val == sum)
{
tmp.push_back(root->val);
ret.push_back(tmp);
return;
}
AC Code：

/**
* 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 > pathSum(TreeNode *root, int sum) {
vector > ret;
vector tmp;
if(root == NULL) return ret;
pathSum_helper(root, sum, tmp, ret);
return ret;
}

private:
void pathSum_helper(TreeNode* root, int sum, vector tmp, vector>& ret)
{
//如果此时的节点就是叶子节点，并且符合sum条件，先push进tmp，再返回。
if(!root->left && !root->right && root->val == sum)
{
tmp.push_back(root->val);
ret.push_back(tmp);
return;
}

if(root != NULL)
{
tmp.push_back(root->val);
int newSum = sum - root->val;
//需要判断是否有左右结点，没有就不递归调用了
if(root->left) pathSum_helper(root->left, newSum, tmp, ret);
if(root->right) pathSum_helper(root->right, newSum, tmp, ret);
}

return;
}
};