PAT-A1020 Tree Traversals 题目内容及题解
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2
题目大意
题目给出某二叉树的后序和中序遍历,求其层序遍历。
解题思路
- 由二叉树后序和中序遍历确定二叉树;
- 对二叉树进行层序遍历并输出,返回0值。
代码
#include<stdio.h>
#include<stdlib.h>
#define maxn 35
typedef struct TreeNode{
int data;
struct TreeNode *lchild,*rchild;
}Node;
int post[maxn],in[maxn],N;
void Init(){
int i;
scanf("%d",&N);
for(i=0;i<N;i++){
scanf("%d",&post[i]);
}
for(i=0;i<N;i++){
scanf("%d",&in[i]);
}
}
Node* Create(postL,postR,inL,inR){
int numL,k=inL;
if(postL>postR){
return NULL;
}
Node* root=(Node*)malloc(sizeof(Node));
root->data=post[postR];
root->lchild=NULL;
root->rchild=NULL;
while(in[k]!=root->data){
k++;
}
numL=k-inL;
root->lchild=Create(postL,postL+numL-1,inL,k-1);
root->rchild=Create(postL+numL,postR-1,k+1,inR);
return root;
}
void Layer(Node* root){
Node *Queue[maxn],*T;
int front=0,rear=0,k=0;
Queue[rear++]=root;
while(front<rear){
T=Queue[front++];
printf("%d",T->data);
if(++k==N){
printf("\n");
}else{
printf(" ");
}
if(T->lchild){
Queue[rear++]=T->lchild;
}
if(T->rchild){
Queue[rear++]=T->rchild;
}
}
return;
}
int main(){
Node* root=NULL;
Init();
root=Create(0,N-1,0,N-1);
Layer(root);
return 0;
}
运行结果