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:

72 3 1 5 7 6 41 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 40int 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;}