1020. Tree Traversals (25) -BFS

题目如下:

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

这是一道很直接的给出中序序列和任一其他序列生成二叉树的问题,本题给出的是后序遍历和中序遍历,利用后序遍历的“左右根”顺序我们知道,后序序列的最后一个元素一定是整棵树的根,从后向前,分别是右、左子树的根,因此通过后序序列可以找到一系列的根,他们的顺序是当前所在的根、右子树的根、左子树的根,每次在中序序列中定位出根的位置,根据中序序列“左根右”的顺序我们知道,根左边的一定是左子树,右边的一定是右子树,就这样递归解决子树问题即可,最后通过BFS来进行层序遍历。


具体实现方法为,设中序序列为inOrder,后序序列为postOrder,设置一个游标变量cur,左右范围变量left、right,cur作为一个全局变量,每次在postOrder中取出一个根,就让cur自减1,首先把拿到的根定位在inOrder中,设根所在的索引为rootIndex,首先建立当前根节点T,然后生成左子树范围left到rootIndex-1和右子树范围rootIndex+1到right,注意由于后序序列倒着走线碰到右子树,因此应该先递归T->right,再递归T->left,当发现left比right大时,说明没有子树,直接返回NULL,当发现left=right时,说明这是一个叶子结点,将结点的left和right置为NULL然后返回,最后一次递归返回时返回的是第一次创建的根结点,也就是整棵树的根,这时便得到了完整的二叉树。


接下来要进行层序遍历,只要对二叉树从根开始调用BFS即可,在结点出队时进行输出。

#include <iostream>
#include <memory.h>
#include <stdio.h>
#include <stdlib.h>
#include <queue>

using namespace std;

#define MAX 40

int postOrder[MAX];
int inOrder[MAX];
int N;
int cur;

typedef struct TreeNode *Tree;
struct TreeNode{

    Tree left;
    Tree right;
    int  num;

};



int findRootIndex(int rootNum){

    for(int i = 0;i < N; i++){
        if(inOrder[i] == rootNum){
            return i;
        }
    }
    return -1;

}

Tree CreateTree(int left, int right){
    if(left > right ) return NULL;
    int root = postOrder[cur];
    cur --;
    int rootIndex = findRootIndex(root);
    Tree T = (Tree)malloc(sizeof(struct TreeNode));
    T->num = root;
    if(left == right){
        T->left = NULL;
        T->right = NULL;
    }else{
        T->right = CreateTree(rootIndex+1,right);
        T->left = CreateTree(left,rootIndex-1);
    }
    return T;

}

void BFS(Tree T){

    bool firstout = true;
    queue<Tree> q;
    q.push(T);

    while(!q.empty()){

        Tree t = q.front();
        q.pop();
        if(firstout){
            firstout = false;
            cout << t->num;
        }else{
            cout << " " << t->num;
        }
        if(t->left != NULL){
            q.push(t->left);
        }
        if(t->right!= NULL){
            q.push(t->right);
        }

    }



}


int main()
{
    cin >> N;
    cur = N-1;
    for(int i = 0; i < N; i++){
        cin >> postOrder[i];
    }
    for(int i = 0; i < N; i++){
        cin >> inOrder[i];
    }

    Tree T = CreateTree(0,cur);

    BFS(T);

    cout << endl;

    return 0;
}


posted on 2015-05-30 13:44  张大大123  阅读(143)  评论(0编辑  收藏  举报

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