1043. Is It a Binary Search Tree (25)

标签：btree   ace   ecif   遍历   case   with   contain   push   nod   

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node‘s key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node‘s key.
• Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line "YES" if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or "NO" if not. Then if the answer is "YES", print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

7
8 6 5 7 10 8 11

Sample Output 1:

YES
5 7 6 8 11 10 8

Sample Input 2:

7
8 10 11 8 6 7 5

Sample Output 2:

YES
11 8 10 7 5 6 8

Sample Input 3:

7
8 6 8 5 10 9 11

Sample Output 3:

NO

解题思路：判断一颗树的前序遍历是否是搜索二叉树或者是其镜像，镜像的意思是：左子树都大于等于根，右子树小于根。做法是建树，判断是否是搜索二叉树，如果不是就判断是否是镜像。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cstdlib>
using namespace std; 
struct Tree{
	int val;
	Tree* left;
	Tree* right;
};
vector<int>value;
bool ind=true;
bool l_num(int a,int b){
	return a<b;
}
bool r_num(int a,int b){
	return a>=b;
}
Tree* bulid(int left,int right,bool cmp(int a,int b)){
	if(!ind)return NULL;//标志已经不能建成树了 
	if(left>right)return NULL;
	Tree* root = new Tree();
	root->val=value[left];
	if(left==right)return root;
	int i;
	for(i=left+1;i<=right;i++){
		if(!cmp(value[i],root->val)){
			break;
		}
	}
	root->left=bulid(left+1,i-1,cmp);
	int j;
	for(j=i;j<=right;j++){
		if(cmp(value[j],root->val)){
			ind=false;
			return NULL;
		}
	}
	root->right=bulid(i,right,cmp);
	return root;
}
int n;
bool tag=true;
void postorder(Tree* root){
	if(root){
		postorder(root->left);
		postorder(root->right);
		if(tag){
			printf("%d",root->val);
			tag=false;
		}else {
			printf(" %d",root->val);
		}
	}
}
int main(){
	scanf("%d",&n);
	int i,j;
	int flag;
	int v;
	for(i=0;i<n;i++){
		scanf("%d",&v);
		value.push_back(v);
	}
	if(n==1){
		printf("YES\n%d\n",value[0]);
		return 0;
	}
	Tree* root = bulid(0,n-1,l_num);
	if(ind){
		printf("YES\n");
		postorder(root);
	}else{
		ind=true;
		root = bulid(0,n-1,r_num);
		if(ind){
			printf("YES\n");
			postorder(root);
		}else{
			printf("NO\n");
		}
	}
}

1043. Is It a Binary Search Tree (25)

标签：btree   ace   ecif   遍历   case   with   contain   push   nod   

原文地址：http://www.cnblogs.com/grglym/p/7765356.html