剑指offer--二叉搜索树的后序遍历序列---AI那点小事

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

思路： 二叉搜索树的性质：根节点大于左子树所有元素，小于右子数的所有元素。后序遍历的话，最后一个元素就为根节点root。若数组为null或者长度为0，则返回true，若数组基本有序（即所有元素已经是升序或降序），返回true，否则，找到划分左右子树的数组元素下标（sequence[i] < root && sequence[i+1] > root），之后判断下标属于[0,i]的子序列是否全小于root，[i,len-2]的子序列是否全大于root。

Java AC代码：

import java.util.Arrays; public class Solution { public boolean VerifyLeftTree(int[] leftsequence , int root){ boolean flag = true; for ( int i = 0 ; i < leftsequence.length ; i++){ if ( leftsequence[i] >= root){ flag = false; break; } } return flag; } public boolean VerifyRightTree(int[] rightsequence , int root){ boolean flag = true; for ( int i = 0 ; i < rightsequence.length ; i++){ if ( rightsequence[i] <= root){ flag = false; break; } } return flag; } //判断sequence是否已经有序 public boolean IsOrder(int [] sequence){ //判断是否已经是升序 int[] tmp = Arrays.copyOfRange(sequence, 0, sequence.length); Arrays.sort(tmp); boolean flag1 = true; for(int i = 0 ; i < sequence.length ; i++){ if ( sequence[i] != tmp[i]){ flag1 = false; break; } } //判断是否已经是降序 int cnt = 0; boolean flag2 = true; for ( int i = tmp.length-1 ; i >= 0 ; i--){ if ( sequence[cnt++] != tmp[i]){ flag2 = false; break; } } boolean flag = flag1 || flag2; return flag; } public boolean VerifySquenceOfBST(int [] sequence) { int partition = 0; int len = sequence.length; if ( sequence == null || len == 0){ return false; } int root = sequence[len-1]; if ( IsOrder(sequence)){ return true; } for ( int i = 0 ; i < len-1 ; i++){ if (sequence[i] < root && sequence[i+1] > root){ partition = i; break; } } int[] left = Arrays.copyOfRange(sequence, 0, partition+1); int[] right = Arrays.copyOfRange(sequence, partition+1, len-1); for ( int i = 0 ; i < left.length ; i++){ System.out.print(left[i]+" "); } System.out.println(); for ( int i = 0 ; i < right.length ; i++){ System.out.print(right[i]+" "); } boolean leftflag = VerifyLeftTree(left,root); boolean rightflag = VerifyRightTree(right,root); boolean flag = leftflag & rightflag; return flag; } }

C++ AC代码

class Solution { public: bool VerifySquenceOfBST(vector<int> sequence) { int len = sequence.size(); return isbst(sequence,0,len-1); } bool isbst(vector<int> sequence,int begin,int end){ if(sequence.empty() || begin > end){ return false; } int root = sequence[end]; int i = begin; for(; i < end ; i++){ if(sequence[i] > root){ //右子树第一个节点 break; } } for(int j = i ; j <end ; j++){ if(sequence[j] < root){ //如果不满足二叉排序树定义，直接返回false return false; } } bool left = true; if(i > begin){ left = isbst(sequence,begin,i-1); } bool right = true; if(end-1 > i){ right = isbst(sequence,i,end-1); } return left&&right; } };

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