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 (≤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

題目大意: 給出樹中每個結點的左右子結點，和一個數據陣列，要求構造出符合的BST(樹的結構要完全對應輸入)。最後輸出該樹的層次遍歷。

 解析:先將資料按小到大排序，觀察知道每個結點的左邊結點總數恰好與對應陣列中的下標一致。所以dfs求出左邊孩子結點數，再用dfs模仿先序遍歷構造出樹，最後用bfs輸出層次遍歷結點。該題只要思路清晰就能解決，本人一次AC過了。

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;

vector<vector<int>> vec;//記錄輸入的左右孩子結點
//arr是原始輸入陣列
//tree對應圖中的樹,下標與資料完全符合
//num是該結點的總孩子結點數，如圖中0結點的孩子總數是8
//lChild該結點的左邊總孩子結點數，如圖中0結點的是5
vector<int> arr,tree,nums,lChild;
int n;

void dfs(int node,int &sum){//計算出該結點的孩子結點數 
	for(int i=0;i<vec[node].size();i++){
		if(vec[node][i] != -1){
			sum++;
			dfs(vec[node][i],sum);
		}		
	}
}
//構造出BST 
void createBST(vector<int> &tree,int node,int low,int high){
	tree[node] = arr[low + lChild[node]];
	if(vec[node][0] != -1)
		createBST(tree,vec[node][0],low,low+lChild[node]-1);
	if(vec[node][1] != -1)
		createBST(tree,vec[node][1],low +lChild[node]+1,high);
}

void bfs(int node){//層次遍歷輸出結果 
	queue<int> que;
	que.push(node);
	int cnt = 0;
	while(!que.empty()){
		int v = que.front();
		que.pop();
		cnt++;
		printf("%d%s",tree[v],cnt!=n?" ":"");
		for(int i=0;i<vec[v].size();i++){
			if(vec[v][i] != -1){
				que.push(vec[v][i]);
			}
		}
	}
}

int main(){
	cin>>n;
	vec.resize(n);
	for(int i=0;i<n;i++){
		int a,b;
		cin>>a>>b;
		vec[i].push_back(a);
		vec[i].push_back(b);
	}		
	arr.resize(n);
	for(int i=0;i<n;i++)
		cin>>arr[i]; 
	sort(arr.begin(),arr.end());//先按從小到大排序 
	nums.resize(n); 
	for(int i=0;i<n;i++){
		int sum = 0;
		dfs(i,sum);
		nums[i] = sum;//獲得每個結點的總孩子結點數 
	}
	lChild.resize(n);
	for(int i=0;i<n;i++){
		int rChild;
		if(vec[i][1] == -1)
			rChild = 0;
		else
			rChild = nums[vec[i][1]] + 1;//先求右邊孩子結點數(記得加上它自身) 
		lChild[i] = nums[i] - rChild;//左邊孩子結點數= 總 - 右邊孩子結點數 
	}
	tree.resize(n);
	createBST(tree,0,0,n-1);
	bfs(0);
		
	return 0;
}