二叉树的重建

#include <iostream>using namespace std;const char preOrder[] = "abdcef";const char inOrder[] = "dbaecf";const int size = sizeof preOrder / sizeof *preOrder;struct Node {char data;Node *left;Node *right;};void Reconstruct(const char *pPre, const char *pIn, int size, Node **pRoot) {if(pPre == NULL || pIn == NULL)return;Node *pTemp = new Node;pTemp->data = *pPre;pTemp->left = NULL;pTemp->right = NULL;if(*pRoot == NULL)*pRoot = pTemp;int pos = 0;const char *ppIn = pIn;while(*ppIn != *pPre) {pos++;ppIn++;}int leftLen = pos;int rightLen = size - pos - 1;if(leftLen > 0)Reconstruct(pPre + 1, pIn, leftLen, &((*pRoot)->left));if(rightLen > 0)Reconstruct(pPre + leftLen + 1, pIn + leftLen + 1, rightLen, &((*pRoot)->right));}void InOrder(Node *root) {if(root) {InOrder(root->left);cout << root->data << " ";InOrder(root->right);}}void PreOrder(Node *root) {if(root) {cout << root->data << " ";PreOrder(root->left);PreOrder(root->right);}}void main() {Node *root = NULL;Reconstruct(preOrder, inOrder, size, &root);PreOrder(root);}

上面是编程之美上的解法，下面是自己的解法：

#include <iostream>#include <queue>using namespace std;struct Node {Node(int i = -1, Node *pLeft = NULL, Node *pRight = NULL) : data(i), left(pLeft), right(pRight) {}int data;Node *left;Node *right;};const char *preOrder = "12473568";const char *inOrder = "47215386";void printLevel(Node *root){if (root == NULL)return;queue<Node *> nqueue;nqueue.push(root);int start = 0;int last = 1;int tempLast = last;while (!nqueue.empty()){Node *top = nqueue.front();nqueue.pop();cout << top->data << " ";if (top->left){nqueue.push(top->left);last++;}if (top->right){nqueue.push(top->right);last++;}start++;if (start == tempLast){tempLast = last;cout << endl;}}}Node *_construct(const char *preOrder, const int preLength, const char *inOrder, const int inLength){if (preOrder == NULL || preLength <= 0 || inOrder == NULL || inLength <= 0)return NULL;int pos = inLength;for (int i = 0; i < inLength; i++){if (inOrder[i] == *preOrder){pos = i;break;}}if (pos == inLength){cout << "wrong" << endl;return NULL;}cout << *preOrder << " ";Node *newNode = new Node(*preOrder - '0');newNode->left = _construct(preOrder + 1, preLength - 1, inOrder, pos);newNode->right = _construct(preOrder + pos + 1, preLength - 1, inOrder + pos + 1, inLength - pos - 1);return newNode;}Node *construct(const char *preOrder, const char *inOrder){if (preOrder == NULL || inOrder == NULL)return NULL;int preLength = strlen(preOrder);int inLength = strlen(inOrder);return _construct(preOrder, preLength, inOrder, inLength);}void preTranverse(Node *root){if (root == NULL)return;cout << root->data << " ";preTranverse(root->left);preTranverse(root->right);}void main(){Node *root = construct(preOrder, inOrder);cout << endl;if (root != NULL){preTranverse(root);}else{cout << "wrong" << endl;}}