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最新推荐文章于 2024-12-18 01:39:59 发布

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本文介绍如何根据给定的二叉树结构和一组不同的整数键唯一地填充二叉搜索树（BST），并输出该树的层次遍历序列。首先定义了二叉搜索树的性质，随后详细描述了输入格式，包括节点数量、子节点关系及键值。通过递归中序遍历算法填充树，并使用队列实现层次遍历来输出结果。

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 (<=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.
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
#include 
using namespace std;
struct Node
{
    int val;
    int lChild;
    int rChild;
    Node() :val(0), lChild(-1), rChild(-1){

    }
};
struct Node node[105];
int num[105];
int cnt = 0;
void inorderTraverse(int root){
    if (root != -1){
        inorderTraverse(node[root].lChild);
        node[root].val = num[cnt++];
        inorderTraverse(node[root].rChild);
    }
}
int main()
{
    int N;
    cin >> N;
    stack s;
    s.push(0);
    int num1, num2,idx;
    for (int i = 0; i < N;i++){
        cin >> num1 >> num2;
        node[i].lChild = num1;
        node[i].rChild = num2;
    }
    for (int i = 0; i < N; i++){
        scanf("%d",&num[i]);
    }
    sort(num,num+N);
    inorderTraverse(0);
    queue q;
    q.push(0);
	int times=1;
    while (!q.empty()){
        idx = q.front();
        q.pop();
        if(times){
        printf("%d",node[idx].val);
    	times=0;
		}
		else {
        	 printf(" %d",node[idx].val);
		}
        if (node[idx].lChild != -1){
            q.push(node[idx].lChild);
        }
        if (node[idx].rChild != -1){
            q.push(node[idx].rChild);
        }
    }
    return 0;
}