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;
}
};