1099 Build A Binary Search Tree (30)

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

题目大意：给定一棵二叉树的结构， 以及一串数字，要求把这些数字填到相应的节点，并且满足二叉树的定义， 输出该二叉树的层序遍历
思路：1.构建二叉树，中序遍历二叉树，确定中序遍历二叉树的节点序列； 
　　  2.对数字排序，因为中序遍历二叉树的得到的是升序的数字， 所以按照中序遍历得到的节点序列，把排序后的数字依次填入节点； 
     3.层序遍历二叉树，输出结果

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
vector<int> v, tree[100], ans(100), in,  level;
void inOrder(int node){
  if(tree[node][0]!=-1)inOrder(tree[node][0]);
  in.push_back(node);
  if(tree[node][1]!=-1) inOrder(tree[node][1]);
}
int main(){
  int n, i;
  scanf("%d", &n);
  v.resize(n);
  for(i=0; i<n; i++){
    int a, b;
    scanf("%d%d", &a, &b);
    tree[i].push_back(a); tree[i].push_back(b);
  }
  for(i=0; i<n; i++) scanf("%d", &v[i]);
  sort(v.begin(), v.end());
  inOrder(0);
  //把节点的值填入节点中
  for(i=0; i<n; i++) ans[in[i]] = v[i];
  queue<int> q;
  q.push(0);
  while(q.size()){//程序遍历
    int tempnode=q.front();
    q.pop();
    level.push_back(ans[tempnode]);
    if(tree[tempnode][0]!=-1) q.push(tree[tempnode][0]);
    if(tree[tempnode][1]!=-1) q.push(tree[tempnode][1]);
  }
  for(i=0; i<n; i++){
    if(i==0) printf("%d", level[i]);
    else printf(" %d", level[i]);
  }
  
  return 0;
}