剑指offer-题24：二叉搜索树的后续遍历序列

题目描述

解决思路：

import java.util.Arrays;

public class Solution {
public boolean VerifySquenceOfBST(int [] sequence) {
if (sequence.length <= 0) {
return false;
}
int rootValue = sequence[sequence.length - 1];
int leftTreeIndex, rightTreeIndex; // 左右子树的索引
for (leftTreeIndex = 0; leftTreeIndex < sequence.length - 1; leftTreeIndex++) {
if (sequence[leftTreeIndex] > rootValue) {
break;
}
}
for (rightTreeIndex = leftTreeIndex; rightTreeIndex < sequence.length - 1; rightTreeIndex++) {
if (sequence[rightTreeIndex] < rootValue) {
return false;
}
}
boolean leftTree = true;
boolean rightTree = true;
if (leftTreeIndex > 0) {
int[] leftArray = Arrays.copyOfRange(sequence, 0, leftTreeIndex);
leftTree = VerifySquenceOfBST(leftArray);
}
if (leftTreeIndex < sequence.length - 1) {
int[] rightArray = Arrays.copyOfRange(sequence, leftTreeIndex, sequence.length - 1);
rightTree = VerifySquenceOfBST(rightArray);
}
return leftTree && rightTree;
}
}

# -*- coding:utf-8 -*-
class Solution:
def VerifySquenceOfBST(self, sequence):
# write code here
if len(sequence) == 0 or sequence is None:
return False
root_value = sequence[-1]
left_tree_index = 0
while sequence[left_tree_index] < root_value and left_tree_index < len(sequence) - 1:
left_tree_index += 1
right_tree_index = left_tree_index
while right_tree_index < len(sequence) - 1:
if sequence[right_tree_index] < root_value:
return False
else:
right_tree_index += 1
right = True
left = True
if left_tree_index > 0:
left = self.VerifySquenceOfBST(sequence[0:left_tree_index])
if left_tree_index < len(sequence) - 1:
right = self.VerifySquenceOfBST(sequence[left_tree_index: -1])
return left and right