[PAT]1099 Build A Binary Search Tree (30 分)(只得22分的原因)
一、题目：
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 Ndistinct 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题目大意是：
给出二叉搜索树的结构，和节点的权值(乱序)，求出二叉搜索树的层次遍历。
解析:
我第一次做事按照：建树、给树节点赋值、层次遍历。这个步骤做的，使用的链表，但是只得22分，看了别人的代码意识到自己的问题所在。如果你也没得满分可能原因如下：
样例输入第一个数n为节点个数，然后跟着n行输入，分别代表第0、1、2、3、、、n-1号节点的左右子节点号。例如：
9 //表示搜索树有9个节点
1 6 //表示第0号节点左右孩子分别为为第1和第6号节点
2 3 //表示第1号节点左右孩子分别为为第2和第3号节点
-1 -1 //表示第2号节点没有左右孩子
-1 4 //表示第3号节点没有左孩子右孩子为第4号节点
5 -1 //表示第4号节点没有右孩子，左孩子为第5号节点
-1 -1 //表示第5号节点没有左右孩子
7 -1 //表示第6号节点左孩子为第7号节点，没有右孩子
-1 8 //表示第7号节点没有左孩子，右孩子为第8号节点
-1 -1 //表示第8号节点没有左右孩子
73 45 11 58 82 25 67 38 42 //表示第0号到第8号节点的值分别为73 45 11....而我第一次做的时候是按照n行输入中非-1值得顺序建树的，我当时使用一个队列第一次入根节点push(0)、然后根据n行输入如果不是-1先入右节点push(right)、再入左节点push(left)。每次都对队列的队首进行赋值，这样做只能得22分。最后按照网上其他人的代码做的才AC。
本题目AC的解题思路为：根据输入建树。对节点值按从小到大排序，所得到的的序列即为BST的中序遍历，据此对节点进行赋值。最后再层次遍历。
下面给出AC代码:
#include <bits/stdc++.h>
using namespace std;
struct node {
int key;
int val;
int leftKey;
int rightKey;
node* left;
node* right;
node(int key, int val=-1, node* left=NULL, node* right=NULL) {
this->key = key;
this->val = val;
this->left = left;
this->right = right;
}
};
node* root;
vector<int> seque;
vector<node*> nodes;
//中序遍历
void inOrder(node* root) {
if(root==NULL) {
return;
}
inOrder(root->left);
root->val = seque[0];
seque.erase(seque.begin());
inOrder(root->right);
}
int main() {
int n;
cin>>n;
nodes.resize(n);
//先申请n个结点，并且记录各个结点的左右孩子节点key
for(int i=0; i<n; i++) {
int left, right;
cin>>left>>right;
node* temp = new node(i);
temp->leftKey = left;
temp->rightKey = right;
nodes[i] = temp;
}
//对n个节点的左右孩子进行赋值
for(int i=0; i<n; i++) {
if(nodes[i]->leftKey != -1) {
nodes[i]->left = nodes[nodes[i]->leftKey];
}
if(nodes[i]->rightKey != -1) {
nodes[i]->right = nodes[nodes[i]->rightKey];
}
}
//现在整个树的结构已经建立，但是节点的val还没有赋值
root = nodes[0];
for(int i=0; i<n; i++) {
int a;
cin>>a;
seque.push_back(a);
}
sort(seque.begin(), seque.end());
//对树中的结点进行赋val
inOrder(root);
//层次遍历
queue<node*> q;
q.push(root);
vector<int > ans;
while(!q.empty()){
ans.push_back(q.front()->val);
if(q.front()->left!=NULL){
q.push(q.front()->left);
}
if(q.front()->right!=NULL){
q.push(q.front()->right);
}
q.pop();
}
for(int i=0; i<ans.size()-1; i++){
cout<<ans[i]<<" ";
}
cout<<ans.back()<<endl;
return 0;
}
