leetcode Path Sum II 关于树的后序遍历

我们利用辅助栈来进行树的后序遍历的时候可以发现栈里保留的是一条根节点到访问的节点的路径，我们可以根据这个性质来解关于树的高度，某条路径和等一系列的问题。

例：

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]
]

代码如下：

/**
* 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<vector<int> > pathSum(TreeNode *root, int sum)
{
vector<vector<int> > result;
if (root == NULL)
{
return result;
}
TreeNode* p = root;
stack<TreeNode*> s;
TreeNode* prev = NULL;
vector<int> tmp;
while (p || !s.empty())
{
if (p)
{
s.push(p);
tmp.push_back(p->val);
p = p->left;
}
else
{
p = s.top();
if (p->right && p->right != prev)
{
p = p->right;
s.push(p);
tmp.push_back(p->val);
p = p->left;
}
else
{
if (p->left == NULL && p->right == NULL)
{
int total = 0;
for (int i = 0; i < tmp.size(); i++)
{
total += tmp[i];
}
if (total == sum)
{
result.push_back(tmp);
}
}
s.pop();
tmp.erase(tmp.end() - 1);
prev = p;
p = NULL;
}
}
}
return result;
}
};