1020 Tree Traversals

题目：

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

 

 

代码：

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
int n;
int pos[32], in[32];
struct Node{
    int value;
    Node* left;
    Node* right;
};
Node* ConstructTree(int posl, int posr, int inl, int inr){
    if(posl > posr){
        return NULL;
    }
    Node* root = new Node;
    root->value = pos[posr];
    int k;
    for(k = inl; in[k] != pos[posr]; k++);
    int numLeft = k - inl;
    root->left = ConstructTree(posl, posl + numLeft - 1, inl, k - 1);
    root->right = ConstructTree(posl + numLeft, posr - 1, k + 1, inr);
    return root;
}
int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &pos[i]);
    }
    for(int i = 0; i < n; i++){
        scanf("%d", &in[i]);
    }
    Node* root = ConstructTree(0, n - 1, 0, n - 1);
    bool flag = false;
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node* tmp = q.front();
        q.pop();
        if(flag == true){
            printf(" ");
        }else{
            flag = true;
        }
        printf("%d", tmp->value);
        if(tmp->left != NULL){
            q.push(tmp->left);
        }
        if(tmp->right != NULL){
            q.push(tmp->right);
        }
    }
    return 0;
}

 

 

总结：

将二叉树以层序遍历输出：

queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node* tmp = q.front();
        q.pop();
        if(flag == true){
            printf(" ");
        }else{
            flag = true;
        }
        printf("%d", tmp->value);
        if(tmp->left != NULL){
            q.push(tmp->left);
        }
        if(tmp->right != NULL){
            q.push(tmp->right);
        }
    }

 根据中序和后序遍历构造二叉树：

Node* ConstructTree(int posl, int posr, int inl, int inr){
    if(posl > posr){
        return NULL;
    }
    Node* root = new Node;
    root->value = pos[posr];
    int k;
    for(k = inl; in[k] != pos[posr]; k++);
    int numLeft = k - inl;
    root->left = ConstructTree(posl, posl + numLeft - 1, inl, k - 1);
    root->right = ConstructTree(posl + numLeft, posr - 1, k + 1, inr);
    return root;
}