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 = 0; i < N; i++)
        scanf("%d%d", &tree[i].left, &tree[i].right);
    Data.resize(N);
    for (int i = 0; i < N; i++)
        scanf("%d", &Data[i]);
    sort(Data.begin(), Data.end());
    getIndex(0, N - 1);
    levelOrder(0);
    return 0;
}
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 != -1)
            Q.push(tree[t].left);
        if (tree[t].right != -1)
            Q.push(tree[t].right);
        if (!Q.empty())
            printf(" ");
    }
}
void getIndex(int t, int M) {
    if (t == -1) {
        return;
    }
    tree[t].rightNum = getNum(tree[t].right);
    tree[t].index = M - tree[t].rightNum;
    getIndex(tree[t].left, tree[t].index - 1);
    getIndex(tree[t].right, M);
}
int getNum(int t) {
    if (t == -1)
        return 0;
    return getNum(tree[t].left) + getNum(tree[t].right) + 1;
}