1099. Build A Binary Search Tree (30)

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.

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

58 25 82 11 38 67 45 73 42

因为给定了结构，因此不能采用一般的方法来创建这棵树，题目指定了采用数组存储，数组存储树的一个问题在于不容易判断树是不是空，以前使用链表存储树的时候，可以通过结点是否为NULL来判断，这里给出一种简单易行的办法。

【数组存储树的方法】

首先建立一个比树的结点个数多1的数组TempT，0位置存储数据-1，然后让树T从位置1开始存储，这样，让空树的索引为-1，当访问T[-1]的时候。实际访问的是TempT[0]，而这里存储的数据正好是-1.因此可以通过这样来防止地址越界和方便的的处理。

【元素插入的处理】

因为二叉搜索树的中序遍历是所有元素的升序，因此将要插入的所有元素进行排序，即可得到树的中序遍历结果。

观察数组索引的增加，是根、左、右的顺序进行的，也就是树的先序遍历，对于二叉搜索树，左侧的元素一定小于右侧，因此只需要设计一个计算左子树和右子树元素个数的函数，然后在中序遍历序列中截取，例如第一次截取，0位置的元素左侧有5个元素，则第6个就是根，后面的就是右子树，然后递归的解决所有子树，完成插入。

【元素的层序遍历问题】

层序遍历和DFS基本是一样的，只需要一个队列，然后进行先序遍历即可。

AC代码如下：

#include <iostream>
using namespace std;

typedef struct TreeNode *Node;
struct TreeNode{
	int left;
	int right;
	int data;
};

Node T;
int *table;

typedef struct Queue_struct *Queue;
struct Queue_struct{
	TreeNode Elements[101];
	int front;
	int rear;
};

Queue createQueue(){

	Queue q = (Queue)malloc(sizeof(Queue_struct));
	q->front = q->rear = 0;
	return q;
}

void EnQueue(Queue q, TreeNode item){
	if (q->rear >= 100)
	{
		return;
	}
	q->Elements[q->rear++] = item;

}

TreeNode DeQueue(Queue q){
	if (q->front == q->rear){
		exit(0);
	}
	return q->Elements[q->front++];
}


void preOrder(TreeNode node){
	
	if (node.left != -2){
		printf("%d ",node.data);
		preOrder(T[node.left]);
		preOrder(T[node.right]);
	}

}

void countNodes(TreeNode node, int* count){

	if (node.left != -2){
		(*count)++;
		countNodes(T[node.left], count);
		countNodes(T[node.right], count);
	}

}

int countChilds(TreeNode node){

	int count = 0;
	countNodes(node, &count);
	return count;
}

int compare(const void* a, const void* b){
	return *(int*)a - *(int*)b;
}

void InsetToBST(int start, int end, Node node){

	if (node->left != -2){
		int begin = start;
		int stop = start + countChilds(T[node ->left]);
		node->data = table[stop];
		InsetToBST(begin, stop - 1, T + node->left);
		begin = stop + 1;
		stop = begin + countChilds(T[node->right]);
		InsetToBST(begin, stop - 1, T + node->right);
	}

}

void MediaOrder(Node T){
	int count = 0;;
	Queue q = createQueue();
	EnQueue(q, T[0]);
	while (q->front != q->rear){
		TreeNode t = DeQueue(q);
		if (count == 0){
			count = 1;
			printf("%d", t.data);
		}
		else{
			printf(" %d", t.data);
		}
		if (T[t.left].data!=-1){
			EnQueue(q, T[t.left]);
		}
		if (T[t.right].data!=-1){
			EnQueue(q, T[t.right]);
		}
	}
	printf("\n");
}

int main(){

	int N;
	cin >> N;
	Node TempTree = (Node)malloc(sizeof(TreeNode)*(N+1));
	TempTree->left = -2;
	TempTree->right = -2;
	TempTree->data = -1;
	T = TempTree + 1;
	for (int i = 0; i < N; i++){
		Node node = T + i;
		node->data = 0;
		scanf("%d%d", &node->left, &node->right);
	}

	table = (int*)malloc(sizeof(int)*N);

	for (int i = 0; i < N; i++){
		scanf("%d", table + i);
	}

	qsort(table, N, sizeof(int), compare);

	InsetToBST(0, N - 1, T + 0);

	MediaOrder(T);
	return 0;

}