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) + ;

 }

巴特西

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

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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