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 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
题目分析
已知二叉查找树的所有非叶子节点的子节点信息，求其层序序列
解题思路
思路 01
- 用二维数组存储所有节点关系，将二叉查找树节点任意序列升序排序即为中序序列
- 建树（左右指针）
- 中序序列打印模拟过程，设置所有节点值信息
- 借助队列，实现层序遍历
思路 02
- 用节点数组建树（类似静态数组）
- 模拟中序序列打印过程，设置所有节点值信息，并记录index（index=i节点的子节点index为2i 1,2i 2（root存放在0下标）），level记录节点所在层
- 用index和level对节点数组排序
- 打印节点数组，即为层序序列
Code
Code 01
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=100;
int n,cnt=0,in[maxn],nds[maxn][2];
struct node {
int data;
node * left=NULL;
node * right=NULL;
node() {}
node(int _data):data(_data) {}
};
node * inOrder(int index) {
if(index==-1) {
return NULL;
}
node * root = new node();
root->left=inOrder(nds[index][0]);
root->data = in[cnt ];
root->right=inOrder(nds[index][1]);
return root;
}
void levelOrder(node * root){
queue<node*> q;
q.push(root);
int index = 0;
while(!q.empty()){
node * now = q.front();
q.pop();
printf("%d",now->data);
if( index<n)printf(" ");
if(now->left!=NULL)q.push(now->left);
if(now->right!=NULL)q.push(now->right);
}
}
int main(int argc,char * argv[]) {
scanf("%d",&n);
int f,r;
for(int i=0; i<n; i ) {
scanf("%d %d",&f, &r);
nds[i][0]=f;
nds[i][1]=r;
}
for(int i=0; i<n; i ) scanf("%d",&in[i]);
sort(in,in n);
node * root = inOrder(0);
levelOrder(root);
return 0;
}Code 02（最优）
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=100;
int n,cnt=0,in[maxn];
struct node {
int data;
int index=-1;
int level=-1;
int left=-1;
int right=-1;
node() {}
node(int _data):data(_data) {}
node(int _left, int _right) {
left=_left;
right=_right;
}
}nds[maxn];
void inOrder(int root, int index, int level) {
if(root==-1) {
return;
}
inOrder(nds[root].left,2*index 1,level 1);
nds[root].index=index;
nds[root].level=level;
nds[root].data=in[cnt ];
inOrder(nds[root].right,2*index 2,level 1);
}
bool cmp(node &n1,node &n2){
if(n1.level!=n2.level)return n1.level<n2.level;
return n1.index<n2.index;
}
int main(int argc,char * argv[]) {
scanf("%d",&n);
int f,r;
for(int i=0; i<n; i ) {
scanf("%d %d",&nds[i].left, &nds[i].right);
}
for(int i=0; i<n; i ) scanf("%d",&in[i]);
sort(in,in n);
inOrder(0,0,0);
sort(nds,nds n,cmp);
for(int i=0;i<n;i ){
if(i!=0)printf(" ");
printf("%d",nds[i].data);
}
return 0;
}