剑指offer--二叉搜索树的后序遍历序列

AI那点小事

发布于 2020-04-18 20:18:14

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发布于 2020-04-18 20:18:14

文章被收录于专栏：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;
        } 
};