剑指Offer算法实现之十九：二叉树的镜像

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题目：请完成一个函数，输入一个二叉树，该函数输出它的镜像。

二叉树节点定义如下：

struct BinaryTreeNode {    int m_nValue;    BinaryTreeNode *m_pLeft;    BinaryTreeNode *m_pRight;};

思路：

遍历二叉树，交换每个节点的左右子树

编译环境：ArchLinux+Clang3.3， C++11

实现一：

#include <iostream>#include <cstring>using namespace std;struct BinaryTreeNode {    int m_nValue;    BinaryTreeNode *m_pLeft;    BinaryTreeNode *m_pRight;};/** 做镜像 **/void do_mirror(BinaryTreeNode *pRoot){    if (!pRoot)        return;    do_mirror(pRoot->m_pLeft);    do_mirror(pRoot->m_pRight);    swap(pRoot->m_pLeft, pRoot->m_pRight);}/**  * 根据二叉树的“单”字符表示形式(形如"1(2,3(,4))")创建二叉树。 * 区间：[start, end)**/BinaryTreeNode *createTree(const char *start, const char *end){    if (start >= end) {        return nullptr;    }    if (end - start == 1) {        return new BinaryTreeNode{*start-'0', nullptr, nullptr};    }    const char *start1 = start+2;    const char *end1 = start+2;    int cnt = 0;    while (true) {        if (*end1 == '(') cnt++;        if (*end1 == ')') cnt--;        if (*end1 == ',' && cnt == 0) break;        end1++;    }    const char *start2 = end1+1;    const char *end2 = end1+1;    cnt = 0;    while (true) {        if (*end2 == '(') cnt++;        if (*end2 == ')' && cnt == 0) break;        if (*end2 == ')') cnt--;        end2++;    }    return new BinaryTreeNode{*start-'0',                     createTree(start1, end1),                    createTree(start2, end2)};}/** 打印二叉树 **/void printTree(BinaryTreeNode *pRoot){    if (!pRoot) return;    cout << pRoot->m_nValue;    if (pRoot->m_pLeft || pRoot->m_pRight) cout << '(';    if (pRoot->m_pLeft) {        printTree(pRoot->m_pLeft);    }    if (pRoot->m_pLeft && pRoot->m_pRight) cout << ',';    if (pRoot->m_pRight) {        printTree(pRoot->m_pRight);    }    if (pRoot->m_pLeft || pRoot->m_pRight) cout << ')';}/** Wrapper **/BinaryTreeNode *createTree(const char *str){    return createTree(str, str+strlen(str));}int main(){    BinaryTreeNode *pRoot = createTree("8(6(5,7),2(9,3))");    printTree(pRoot); cout << endl;    do_mirror(pRoot);    printTree(pRoot); cout << endl;}