5-1 Build A Binary Search Tree (30分)

5-1 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 ($\le$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 formatleft_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

#include
#include
#include

struct Tree{
	int num;
	int left;
	int right;
};

int main(void)
{
	int n, m, temp, x, top, k, number;
	scanf("%d",&n);
	struct Tree a[n], c[n];
	struct Tree T;
	int b[n], d[n];
	int front, rear;
	for(int i=0; ib[j+1])
			{
				temp=b[j];
				b[j]=b[j+1];
				b[j+1]=temp;
			}
		}
	}
	number=0;
	top=0;
	k=0;
	while(top!=0||number!=-1)
	{
	    while(number!=-1)
	    {
		    d[top++]=number;
		    number=a[number].left;
	    }
	    number=d[--top];
	    a[number].num=b[k++];
	    if(number!=-1)
	    {
	    	number=a[number].right;
		}
	}
	number=0;
	c[0]=a[0];
	front=0;
	rear=1;
	while(front!=rear)
	{
		number=c[front].left;
		if(number!=-1)
		{
			c[rear++]=a[number];
		    
	    }
		number=c[front++].right;
		if(number!=-1)
		{
		    c[rear++]=a[number];
	    }
	}
	for(int i=0; i