剑指offer2-重建二叉树

来源：互联网发布：套路网络用语编辑：程序博客网时间：2024/06/06 14:02

二叉树的遍历：
先序遍历：先访问根节点，再访问左子树，最后访问右子树；
中序遍历：先访问左子树，再访问根节点，最后访问右子树；
后序遍历：先访问左子数，再访问右子树，最后访问根节点；
一般给定一棵二叉树的中序遍历和先序遍历或者给定一个二叉树的中序遍历和后序遍历，这棵二叉树就可以确定下来。仅仅给定先序遍历和后序遍历是无法确定的。
题目：
给定某二叉树的先序和中序遍历结果，请确定该二叉树：

#include<iostream>using namespace std;struct TreeNode{    int value;    TreeNode* leftTree;    TreeNode* rightTree;};TreeNode* constructCore(int* startPreOrder, int* endPreOrder, int* startInOrder,                        int* endInOrder){    int rootValue = startPreOrder[0];    TreeNode* root = new TreeNode;    root->value = rootValue;    root->leftTree = root->rightTree = NULL;    if (startPreOrder == endPreOrder)    {        if (startInOrder == endInOrder && *startPreOrder == *endPreOrder)        {            return root;//只有一个元素时        }        else        {            throw exception("Invalid input.");        }    }    int* rootInOrder = startInOrder;    while (rootInOrder <= endInOrder && *rootInOrder != rootValue)    {        rootInOrder++;    }    if (rootInOrder == endInOrder && *rootInOrder != rootValue)    {        throw exception("Invalid inout.");    }    int leftLength = rootInOrder - startInOrder;    int rightLength = endPreOrder - startPreOrder - leftLength;    if (leftLength > 0)    {        //构建左子树        root->leftTree = constructCore(startPreOrder + 1, startPreOrder + leftLength,                                       startInOrder, rootInOrder - 1);    }    if (rightLength > 0)    {        root->rightTree = constructCore(startPreOrder + leftLength + 1, endPreOrder,                                        rootInOrder + 1, endInOrder);    }    return root;}TreeNode* construct(int* preOrder, int* inOrder, int length){    if (preOrder == NULL || inOrder == NULL || length <= 0)    {        return NULL;    }    return constructCore(preOrder, preOrder + length - 1, inOrder,                         inOrder + length - 1);}void printInOrder(TreeNode* root){    if (root->leftTree != NULL)    {        printInOrder(root->leftTree);    }    cout << root->value << " ";    if (root->rightTree != NULL)    {        printInOrder(root->rightTree);    }}void printPreOrder(TreeNode* root){    cout << root->value << " ";    if (root->leftTree != NULL)    {        printPreOrder(root->leftTree);    }    if (root->rightTree != NULL)    {        printPreOrder(root->rightTree);    }}void printPosterOrder(TreeNode* root){    if (root->leftTree != NULL)    {        printPosterOrder(root->leftTree);    }    if (root->rightTree != NULL)    {        printPosterOrder(root->rightTree);    }    cout << root->value << " ";}int main(){    int preOrder[] = {1, 2, 4, 7, 3, 5, 6, 8};    int inOrder[] = {4, 7, 2, 1, 5, 3, 8, 6};    int length = sizeof(preOrder) / sizeof(int);    TreeNode* root = construct(preOrder, inOrder, length);    cout << "中序遍历：" << endl;    printInOrder(root);    cout << endl;    cout << "先序遍历" << endl;    printPreOrder(root);    cout << endl;    cout << "后序遍历" << endl;    printPosterOrder(root);    cout << endl;    return 0;}

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