CS-Notes/notes/33. 二叉搜索树的后序遍历序列.md
2020-11-17 00:32:18 +08:00

35 lines
1.3 KiB
Java
Raw Permalink Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

# 33. 二叉搜索树的后序遍历序列
[NowCoder](https://www.nowcoder.com/practice/a861533d45854474ac791d90e447bafd?tpId=13&tqId=11176&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
## 题目描述
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的结果假设输入的数组的任意两个数字都互不相同
例如下图是后序遍历序列 1,3,2 所对应的二叉搜索树
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/13454fa1-23a8-4578-9663-2b13a6af564a.jpg" width="150"/> </div><br>
## 解题思路
```java
public boolean VerifySquenceOfBST(int[] sequence) {
if (sequence == null || sequence.length == 0)
return false;
return verify(sequence, 0, sequence.length - 1);
}
private boolean verify(int[] sequence, int first, int last) {
if (last - first <= 1)
return true;
int rootVal = sequence[last];
int cutIndex = first;
while (cutIndex < last && sequence[cutIndex] <= rootVal)
cutIndex++;
for (int i = cutIndex; i < last; i++)
if (sequence[i] < rootVal)
return false;
return verify(sequence, first, cutIndex - 1) && verify(sequence, cutIndex, last - 1);
}
```