由后序遍历和中序遍历构建二叉树（C++语言）

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设计思路：

char* post为后序遍历的顺序
char* in为中序遍历的顺序
首先建立一个指针p，用循环在in中找到根节点
left为左子树个数=p-in（指针差值）
right为右子树个数（n-left-1）
之后递归调用该函数构建左右子树

注意：要想构建二叉树，必须知道中序遍历，这样才可以知道根节点，进而确定左右子树
有前序和后序不能够构建二叉树

代码：

/**
 *作者：魏宝航
 *2020年11月27日,下午21:58
 */
#include<iostream>
using namespace std;
class Node {
public:
    char ch;
    Node* left;
    Node* right;
    Node() {
        ch = '\0';
        left = NULL;
        right = NULL;
    }
    Node(char ch, Node* left, Node* right) {
        this->ch = ch;
        this->left = left;
        this->right = right;
    }
};
static int i = 0;
//#号法创建二叉树
Node* CreateTree(vector<char> arr,Node* root) {
    if (i < arr.size()) {
        char temp = arr[i++];
        if (temp == '#') {
            return NULL;
        }
        else {
            root = new Node();
            root->ch = temp;
            root->left = CreateTree(arr, root->left);
            root->right = CreateTree(arr, root->right);
        }
    }
    return root;
}
//二叉树的前序遍历
void preOrder(Node* root) {
    if (root == NULL) {
        return;
    }
    cout << root->ch << " ";
    preOrder(root->left);
    preOrder(root->right);
}
//二叉树的中序遍历
void midOrder(Node* root) {
    if (root == NULL) {
        return;
    }
    midOrder(root->left);
    cout << root->ch << " ";
    midOrder(root->right);
}
//二叉树的后序遍历
void postOrder(Node* root) {
    if (root == NULL) {
        return;
    }
    postOrder(root->left);
    postOrder(root->right);
    cout << root->ch << " ";
}
//计算二叉树的高度
int treeDepth(Node* root) {
    int left = 0, right = 0;
    if (root == NULL) {
        return 0;
    }
    else {
        left = treeDepth(root->left) + 1;
        right = treeDepth(root->right) + 1;
        return max(left, right);
    }
}
Node* buildTree1(char* pre, char* in, int n) {
    if (n == 0) {
        return NULL;
    }
    Node* root = new Node();
    root->ch = *pre;
    char* p;
    for (p = in; p < in + n; p++) {
        if (*p == *pre) {
            break;
        }
    }
    int left = p - in;//左子树个数
    int right = n - left - 1;//右子树个数
    root->left = buildTree1(pre + 1, in, left);
    root->right = buildTree1(pre + left + 1, p + 1, right);
    return root;
}
Node* buildTree2(char* post, char* in, int n) {
    if (n == 0) {
        return NULL;
    }
    Node* root = new Node();
    root->ch = *(post + n-1);
    char* p;
    for (p = in; p < in + n; p++) {
        if (*p == *(post + n - 1)) {
            break;
        }
    }
    int left = p - in;
    int right = n - left - 1;
    root->left = buildTree2(post,in, left);
    root->right = buildTree2(post + left , p + 1, right);
    return root;
}
int main() {
    char pre[] = { 'A','B','D','E','C','F' };
    char mid[] = { 'D','B','E','A','C','F' };
    char post[]= { 'D','E','B','F','C','A' };
    Node* root = new Node();
    root = buildTree2(post, mid, 6);
    cout << "前序遍历：";
    preOrder(root);
    cout << endl;
    cout << "中序遍历：";
    midOrder(root);
    cout << endl;
    cout << "后序遍历：";
    postOrder(root);
    cout << endl;
}