PAT Advanced 1099 Build A Binary Search Tree (30) [⼆叉查找树BST]

题目

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.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format left_index right_index, provided that the nodes are numbered from 0 to N−1, and 0 is always the root. If one child is missing, then −1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.
Output Specification:
For each test case, print in one line the level order traversal sequence of that tree. All the numbers must
be separated by a space, with no extra space at the end of the line.
Sample Input:
9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42
Sample Output:
58 25 82 11 38 67 45 73 42

题目分析

已知二叉查找树的所有非叶子节点的子节点信息，求其层序序列

解题思路

思路 01

1. 用二维数组存储所有节点关系，将二叉查找树节点任意序列升序排序即为中序序列
2. 建树（左右指针）
3. 中序序列打印模拟过程，设置所有节点值信息
4. 借助队列，实现层序遍历

思路 02

1. 用节点数组建树（类似静态数组）
2. 模拟中序序列打印过程，设置所有节点值信息，并记录index（index=i节点的子节点index为2i+1,2i+2（root存放在0下标）），level记录节点所在层
3. 用index和level对节点数组排序
4. 打印节点数组，即为层序序列

Code

Code 01

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=100;
int n,cnt=0,in[maxn],nds[maxn][2];
struct node {
	int data;
	node * left=NULL;
	node * right=NULL;
	node() {}
	node(int _data):data(_data) {}
};
node * inOrder(int index) {
	if(index==-1) {
		return NULL;
	}
	node * root = new node();
	root->left=inOrder(nds[index][0]);
	root->data = in[cnt++];
	root->right=inOrder(nds[index][1]);
	return root;
}
void levelOrder(node * root){
	queue<node*> q;
	q.push(root);
	int index = 0;
	while(!q.empty()){
		node * now = q.front();
		q.pop();
		printf("%d",now->data);
		if(++index<n)printf(" "); 
		if(now->left!=NULL)q.push(now->left);
		if(now->right!=NULL)q.push(now->right);
	}
}
int main(int argc,char * argv[]) {
	scanf("%d",&n);
	int f,r;
	for(int i=0; i<n; i++) {
		scanf("%d %d",&f, &r);
		nds[i][0]=f;
		nds[i][1]=r;
	}
	for(int i=0; i<n; i++) scanf("%d",&in[i]);
	sort(in,in+n);
	node * root = inOrder(0);
	levelOrder(root);
	return 0;
}

Code 02（最优）

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=100;
int n,cnt=0,in[maxn];
struct node {
	int data;
	double index=-1; //最坏情况，每一层一个节点，index最大为2^100 ，测试使用long long也有两个测试点不通过
	int left=-1;
	int right=-1;
	node() {}
	node(int _data):data(_data) {}
	node(int _left, int _right) {
		left=_left;
		right=_right;
	}
}nds[maxn];
void inOrder(int root, double index) {
	if(root==-1) {
		return;
	}
	inOrder(nds[root].left,2*index+1);
	nds[root].index=index;
	nds[root].data=in[cnt++];
	inOrder(nds[root].right,2*index+2);
}
bool cmp(node &n1,node &n2){
	return n1.index<n2.index;
}
int main(int argc,char * argv[]) {
	scanf("%d",&n);
	int f,r;
	for(int i=0; i<n; i++) {
		scanf("%d %d",&nds[i].left, &nds[i].right);
	}
	for(int i=0; i<n; i++) scanf("%d",&in[i]);
	sort(in,in+n);
	inOrder(0,0);
	sort(nds,nds+n,cmp);
	for(int i=0;i<n;i++){
		if(i!=0)printf(" ");
		printf("%d",nds[i].data);
	}
	return 0;
}