pat 甲级 1099. Build A Binary Search Tree (30)
1099. Build A Binary Search Tree (30)
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
题意:给定n个数插入BST,并且插入数值后的BST结构已经确定,依据这个结构来推断BST上每个节点对应的数值,并且层序遍历BST。
思路:一开始的思路比较复杂,将n各数值从小到大排列,我先搜索出了BST上每个节点左右子树的节点个数。在此基础上就可以递推的确定每个节点对应的数值在数列上的位置。假设
已经确定某一个节点x对应的数值在数列上的位置pos,那么其节点x的左儿子的数值所在位置与pos的距离间隔就是左儿子的右子树节点个数(原因:比左儿子的数值大又比x的数值小,这些节点的数值当然都存储在左儿子的右子树上),从而可以推断左儿子的数值。右儿子的数值推断方法类似。
但其实不需要如此复杂,按照中序遍历的顺序就可以直接推断出各个节点上对应的数值。。。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<set> #include<queue> using namespace std; #define INF 0x3f3f3f #define N_MAX 100+5 int n; struct Node { int l_child, r_child; int key; }node[N_MAX]; vector<int>vec; pair<int, int>num[N_MAX]; /* int dfs(int x) {//查询每个节点左右儿子的数量 int left_num=0, right_num=0; if(node[x].l_child!=-1)left_num = dfs(node[x].l_child); if (node[x].r_child != -1)right_num = dfs(node[x].r_child); num[x] = make_pair(left_num, right_num); return right_num + left_num+1; } void dfs2(int x,int pos) {//节点x上的数值为vec[pos],考虑节点x与儿子节点的距离来推断儿子节点的位置 node[x].key = vec[pos]; int pos_left = pos - num[node[x].l_child].second - 1; int pos_right = pos + num[node[x].r_child].first + 1; if(node[x].l_child!=-1)dfs2(node[x].l_child, pos_left); if(node[x].r_child!=-1)dfs2(node[x].r_child, pos_right); }*/ int step = 0; void inorder(int x) {//其实按照中序遍历的顺序就可以依次确定每个节点的数值key if(node[x].l_child!=-1)inorder(node[x].l_child); node[x].key = vec[step++]; if(node[x].r_child!=-1)inorder(node[x].r_child); } int output[N_MAX]; void bfs(int root) { queue<int>que; que.push(root); int cnt = 0; while(!que.empty()) { int x = que.front(); que.pop(); output[cnt++] = node[x].key; if(node[x].l_child!=-1)que.push(node[x].l_child); if(node[x].r_child!=-1)que.push(node[x].r_child); } } int main(){ while (scanf("%d",&n)!=EOF) { for (int i = 0; i < n;i++) { int l, r; scanf("%d%d",&l,&r); node[i].l_child = l, node[i].r_child = r; } vec.resize(n); for (int i = 0; i < n; i++)scanf("%d",&vec[i]); sort(vec.begin(),vec.end()); //dfs(0); //dfs2(0, num[0].first); inorder(0); bfs(0); for (int i = 0; i < n; i++)printf("%d%c",output[i],i+1==n?'\n':' '); } return 0; }