最近在刷PAT题，将自己的解题过程和代码写在blog上。
Tree Traversals (25)
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 7Sample Output:
4 1 6 3 5 7 2
这题自己想了挺久，根据后序排列和中序排列的数组来确定一棵二叉树 。解题思路：根据后序排列可以确定树的根，根据中序排列可以比对根节点确定左右子树。然后因为后序排列安照 左-》右-》根的顺序遍历，而中序排列按照左->根->右的顺序，可以
确定后序排列中的从左到右直到达到根节点左边的肯定为左子树；即不断确定根节点，左右子树来完善这颗树。
当树行成后需要按层级打印，这里注意不是按照先序遍历的方式打印，这时需要用到队列，将根元素先入队，再将左右子树节点入队，根据节点不为空的原则进行出队操作，一直到队列为空；
1 // 2 // main.cpp 3 // 1020 4 // 5 // Created by apple on 14-2-13. 6 // Copyright (c) 2014年 apple. All rights reserved. 7 // 8 9 #include <iostream> 10 #include <queue> 11 using namespace std; 12 #define MAX 31 13 int inorder[MAX]; 14 int postorder[MAX]; 15 typedef struct Node{ 16 int value; 17 struct Node *lchild; 18 struct Node *rchlid; 19 }Node; 20 21 void LevelTravel(Node *root); 22 Node *createtree(int istart,int iend,int pstart,int pend) 23 { 24 int i,j; 25 Node *node = new Node(); 26 27 node->lchild = NULL; 28 node->rchlid = NULL; 29 node->value = postorder[pend]; 30 if (istart>iend || pstart>pend) 31 { 32 return NULL; 33 } 34 35 36 for(i = istart,j = pstart; i < iend && j < pend; i++,j++) 37 { 38 if (inorder[i] == postorder[pend]) 39 { 40 break; 41 } 42 } 43 44 45 node->lchild = createtree(istart,i-1,pstart,j-1); 46 node->rchlid = createtree(i+1,iend,j,pend-1); 47 48 49 return node; 50 } 51 52 53 54 int main() 55 { 56 int i; 57 int num; 58 cin>>num; 59 for (i = 0; i<num; i++) 60 { 61 cin>>postorder[i]; 62 } 63 64 for (i = 0; i<num; i++) 65 { 66 cin>>inorder[i]; 67 } 68 69 Node *root = NULL; 70 root = createtree(0,num-1,0,num-1); 71 72 LevelTravel(root); 73 74 return 0; 75 } 76 77 void LevelTravel(Node *root) 78 { 79 queue<Node*>q; 80 if (root!=NULL) 81 { 82 printf("%d",root->value); 83 if(root->lchild!=NULL)q.push(root->lchild); 84 if(root->rchlid!=NULL)q.push(root->rchlid); 85 } 86 else 87 return; 88 while(!q.empty()) 89 { 90 Node *node = q.front(); 91 92 if (node == NULL) 93 { 94 continue; 95 } 96 q.pop(); 97 printf(" %d",node->value); 98 99 if(node->lchild!=NULL) q.push(node->lchild); 100 101 if(node->rchlid!=NULL) q.push(node->rchlid); 102 103 } 104 printf(" "); 105 106 }