PAT 1099. Build A Binary Search Tree (30)

1099. Build A Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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
  

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提交代码

深搜进行赋值，广搜进行层序遍历

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int lch[maxn], rch[maxn], a[maxn], val[maxn], k;

void dfs(int now)
{
	if(now < 0) return ;
	dfs(lch[now]);
	val[now] = a[k++];
	dfs(rch[now]);
}

void bfs(int root)
{
	queue<int> q;
	q.push(root);
	int flag = 0;
	while(q.size())
	{
		root = q.front();
		q.pop();
		if(flag++) printf(" ");
		printf("%d", val[root]);
		if(lch[root] != -1) q.push(lch[root]);
		if(rch[root] != -1) q.push(rch[root]);
	}
}

int main(void)
{
	int n;
	while(cin >> n)
	{
		for(int i = 0; i < n; i++) scanf("%d%d", &lch[i], &rch[i]);
		for(int i = 0; i < n; i++) scanf("%d", &a[i]);
		sort(a, a+n);
		k = 0, dfs(0), bfs(0);
		printf("\n");
	}
	return 0;
}