带你快速刷完67道剑指offer

No23、二叉搜索树的后序遍历序列

牛客网原题链接

题目描述

从上往下打印出二叉树的每个节点，同层节点从左至右打印。

示例1

输入

{5,4,#,3,#,2,#,1}

返回值

[5,4,3,2,1]

1、递归写法，树主要的做法就是递归

bool VerifySquenceOfBST(vector<int> sequence) {
    if (sequence.empty())  return false;
    if (sequence.size() == 1) return true;
    return VerifySquenceOfBSTCore(sequence, 0, sequence.size()-1);
}

bool VerifySquenceOfBSTCore(vector<int>& sequence, int start, int end) {
    if (start >= end) return true;
    int low = start;
    while (low < end && sequence[low] < sequence[end])  ++low;

    for (int i = low; i < end; ++i) {
        if (sequence[i] <= sequence[end]) return false;
    }

    return  VerifySquenceOfBSTCore(sequence, start,low-1) &&
        VerifySquenceOfBSTCore(sequence, low,end-1);
}

二刷：依然没有思路，值得再看一遍

1、并没有想象中的难，下次应该仔细想一想的

    bool VerifySquenceOfBSTCore(vector<int>&sequence,int low,int high){
        if(low >= high) return true;
        int start = low;
        while(start < high && sequence[start] < sequence[high]) ++start;//二叉搜索树，左右根，左子树全部小于根
        //右子树全部打大于根，找到第一个大于根的元素，那么在他之前都是左子树，之后都是右子树
        for(int i = start;i < high; ++i)
            if(sequence[i] <= sequence[high]) return false; //右子树必须全部大于根，否则就是假
        return VerifySquenceOfBSTCore(sequence, low, start-1) //判断当前节点的其左子树
            && VerifySquenceOfBSTCore(sequence, start, high-1);//判断当前节点的其右子树

    }
    bool VerifySquenceOfBST(vector<int> sequence) {

        if(sequence.empty()) return false;//为空，则为假
        if(sequence.size() == 1) return true;//只有一个元素，为真
        
        return VerifySquenceOfBSTCore(sequence,0,sequence.size()-1);
    }