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:

91 62 3-1 -1-1 45 -1-1 -17 -1-1 8-1 -173 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;}