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# 面试题33
输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历结果。如果是则返回 true，否则返回 false。假设输入的数组的任意两个数字都互不相同。
```
参考以下这颗二叉搜索树：
5
/ \
2 6
/ \
1 3
示例 1：
输入: [1,6,3,2,5]
输出: false
示例 2：
输入: [1,3,2,6,5]
输出: true
```
## 提示：
* 数组长度 <= 1000
## Solutions
1. **mono stack**
2. reference:
3. This solution is based on the fact that postorder traversal is the opposite of preorder traversal. ie: preorder(Similar): `root right left` -> postorder: `left right root`
4. Maintain a monotonically increasing stack which records the path when traversing the right subtree(Mono increasing).
5. This solution works only if there are no duplicate elements in the given sequence.
```cpp
class Solution {
public:
bool verifyPostorder(vector& postorder) {
stack s;
int cur = postorder.size() - 1;
int rootval = INT_MAX;
for (int cur = postorder.size() - 1; cur >= 0; cur--) {
if (postorder[cur] > rootval)
return false;
// find the root node
while (!s.empty() && s.top() > postorder[cur]) {
rootval = s.top();
s.pop();
}
s.push(postorder[cur]);
}
return true;
}
};
```
1. **recursion**
2. In each recursive call, find the first element of the right subtree(val > root->val).
```cpp
class Solution {
public:
bool valid(vector & postorder, int lo, int hi) {
if (lo >= hi)
return true;
// the root val is the last element
int rootval = postorder[hi];
int mid = lo;
while (postorder[mid] < rootval && mid < hi)
mid++;
// check if the right tree is valid
for (int r = mid; r < hi; r++)
if (postorder[r] < rootval)
return false;
return valid(postorder, lo, mid - 1) && valid(postorder, mid, hi - 1);
}
bool verifyPostorder(vector& postorder) {
return valid(postorder, 0, postorder.size() - 1);
}
};
```