剑指Offer-- 二叉搜索树的后序遍历序列判断

版本1：C++

class Solution {
public:
    bool VerifySquenceOfBST(vector<int> sequence) {
        if (sequence.size() == 0)
            return false;
        int len = sequence.size(),i = 0,j;
        int root = sequence[len - 1];         // 根节点
        vector<int> leftTree, rightTree;
        while (sequence[i] < root){           // 找出左子树的范围 0 ~ i-1
            leftTree.push_back(sequence[i]);  // 将左子树存入一个vector
            i++;
        }   
        for (j = i; j < len - 1; j++){        // 在右子树的范围 j ~ len-1 内判断有没有不合理的数字
            if (sequence[j] < root){
                return false;
            }
            rightTree.push_back(sequence[j]);  // 将右子树存入一个vector
        }
        bool resofleft = true, resofright = true;             // 这里的初始化重要，是控制判断为真情况的
        if (leftTree.size() > 0){                             // 保证子树为空时能够跳过判断
            resofleft = VerifySquenceOfBST(leftTree);         // 递归判断左子树
        }
        if (rightTree.size() > 0){
            resofright = VerifySquenceOfBST(rightTree);       // 递归判断右子树
        }
        return (resofleft && resofright);                     // 左右子树都返回为真时，才为真
    }
};

 版本2：python

# -*- coding:utf-8 -*-
class Solution:
    def VerifySquenceOfBST(self, sequence):
        # write code here
        if not sequence:
            return False
        leng = len(sequence)
        root = sequence[leng - 1]
        i = 0
        ltree = []
        rtree = []
        while sequence[i] < root:
            ltree.append(sequence[i])
            i = i + 1
        j = i
        for j in range(j, leng - 1):
            if sequence[j] < root:
                return False
            rtree.append(sequence[j])
        
        leftTree = True
        rightTree = True
        if len(ltree) :
            leftTree = self.VerifySquenceOfBST( ltree)
        if len(rtree) :
            rightTree = self.VerifySquenceOfBST( rtree)
            
        return (leftTree and rightTree)