剑指offer--面试题24: 二叉搜索树的后序遍历序列

来源：互联网发布：网络工程和网络布线编辑：程序博客网时间：2024/06/06 03:02

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。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。

python实现：

# -*- coding:utf-8 -*-class Solution:    def VerifySquenceOfBST1(self, sequence):        # write code here        #二叉搜索树的特性        n = len(sequence)        if n==0:            return False        if n==1:            return True        if sequence[0]>sequence[-1]:#没有左子树            for i in range(1, n-1):                if sequence[i]<sequence[-1]:                    return False            return self.VerifySquenceOfBST(sequence[:-1])                 rightStart = n-1                 for i in range(1, n-1):            if sequence[i]>sequence[-1]:                rightStart = i            else:                if rightStart<n-1:                    return False                     if rightStart==n-1:#没有右子树            return self.VerifySquenceOfBST(sequence[:-1])        else:            return self.VerifySquenceOfBST(sequence[:rightStart]) and self.VerifySquenceOfBST(sequence[rightStart:-1])

    #写法2：    def VerifySquenceOfBST(self, sequence):        n = len(sequence)        if n==0:            return False        if n==1:            return True        root = sequence[-1]        leftEnd = -2        rightStart = n-1        for i in range(n-1):            if sequence[i]>root:                if rightStart==n-1:                    rightStart = i                    leftEnd = i-1            else:                if leftEnd!=-2:                    return False        left = True        if leftEnd>=0:#有左子树            left = self.VerifySquenceOfBST(sequence[:leftEnd+1])        right = True        if rightStart<n-1:#有右子树            right = self.VerifySquenceOfBST(sequence[rightStart:-1])        return left and right

c++实现：

class Solution {public:    bool VerifySquenceOfBST(vector<int> sequence) {        return verify(sequence, 0, sequence.size()-1);    }         bool verify(vector<int> &sequence, int start, int end){        if((end-start+1)==0)            return false;        int root=sequence[end];        int leftEnd=start-2, rightStart=end;        for(int i=start;i<end;i++){             if(sequence[i]<root){                if(leftEnd!=start-2)//已经找到左子树终点，现在又遇到比跟小的数                    return false;            }else{                if(rightStart==end){//还没找到右子树                    rightStart = i;                    leftEnd=i-1;//rightStart=0时leftEnd=-1,这也是为什么leftEnd初始化为2的原因                }            }        }        bool left=true;        if(leftEnd>=start)//有左子树            left = verify(sequence, start, leftEnd);        bool right=true;        if(rightStart<end)//有右子树            right = verify(sequence, rightStart, end-1);        return (left && right);    }};

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