PAT甲级——1099 Build A Binary Search Tree (二叉搜索树)

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1099 Build A Binary Search Tree (30 分)

 

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 (≤) 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 − will represent the NULL child pointer. Finally Ndistinct 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

题目大意：将N个数放入一棵定型了的二叉树，使其满足二叉搜索树的性质。

思路：先将数据Data排好序，二叉树中存放数据的下标就行。

对于BST中的每个节点，它的key值对应的下标 index = 其上层节点传递过来的 M - 其右子树节点的个数 rightNum。若当前节点是其parent节点的左孩子，这个传递过来的M值就是parent节点的下标；若当前节点是parent节点的右孩子，那么M就是其parent节点的M。根节点的M值为N-1。

 #include <iostream>
 #include <vector>
 #include <queue>
 #include <algorithm>
 using namespace std;
 struct node {
     int left, right,
         rightNum,
         index;
 };
 vector <node> tree;
 vector <int> Data;
 int getNum(int t);
 void getIndex(int t, int M);
 void levelOrder(int t);
 int main()
 {
     int N;
     scanf("%d", &N);
     tree.resize(N);
     for (int i = ; i < N; i++)
         scanf("%d%d", &tree[i].left, &tree[i].right);
     Data.resize(N);
     for (int i = ; i < N; i++)
         scanf("%d", &Data[i]);
     sort(Data.begin(), Data.end());
     getIndex(, N - );
     levelOrder();
     return ;
 }
 void levelOrder(int t) {
     queue <int> Q;
     Q.push(t);
     while (!Q.empty()) {
         t = Q.front();
         Q.pop();
         printf("%d", Data[tree[t].index]);
         if (tree[t].left != -)
             Q.push(tree[t].left);
         if (tree[t].right != -)
             Q.push(tree[t].right);
         if (!Q.empty())
             printf(" ");
     }
 }
 void getIndex(int t, int M) {
     if (t == -) {
         return;
     }
     tree[t].rightNum = getNum(tree[t].right);
     tree[t].index = M - tree[t].rightNum;
     getIndex(tree[t].left, tree[t].index - );
     getIndex(tree[t].right, M);
 }
 int getNum(int t) {
     if (t == -)
         return ;
     return getNum(tree[t].left) + getNum(tree[t].right) + ;
 }