[编程之美-14]求二叉树中节点的最大距离

题目:如果我们把二叉树看成一个图，父子节点之间的连线看成是双向的，我们姑且定义”距离”为两节点之间边的个数。写一个程序，求一棵二叉树中相距最远的两个节点之间的距离. 

例如: 

10 

/ \ 

5 12 

/ \ 

4 7 

这棵树的话,最大距离为3.分别路径为4,5,10,12共3条边,7,5,10,12共3条边,所以最大距离为3.

递归的思想,分别考虑左右子树的,从根节点开始.

代码如下:

#include<iostream>
using namespace std;

struct BiTreeNode
{
int m_nValue;
BiTreeNode *m_pleft;
BiTreeNode *m_pright;
int m_nMaxLeft;
int m_nMaxRight;
};

int nMaxLen = 0;

void findMaxLen(BiTreeNode *pRoot);

void addBiTreeNode(BiTreeNode *&pCurrent, int value);

int main()
{
BiTreeNode *pRoot = NULL;
addBiTreeNode(pRoot, 10);
addBiTreeNode(pRoot, 5);
addBiTreeNode(pRoot, 4);
addBiTreeNode(pRoot, 7);
addBiTreeNode(pRoot, 12);

findMaxLen(pRoot);
cout<<nMaxLen<<endl;
return 0;
}

void findMaxLen(BiTreeNode *pRoot)
{
if(pRoot == NULL)
return ;

if(pRoot->m_pleft == NULL)
pRoot->m_nMaxLeft = 0;

if(pRoot->m_pright == NULL)
pRoot->m_nMaxRight = 0;

if(pRoot->m_pleft != NULL)
findMaxLen(pRoot->m_pleft);

if(pRoot->m_pright != NULL)
findMaxLen(pRoot->m_pright);

if(pRoot->m_pleft != NULL)
{
int nTempMax = 0;

if(pRoot->m_pleft->m_nMaxLeft > pRoot->m_pleft->m_nMaxRight)
nTempMax = pRoot->m_pleft->m_nMaxLeft;
else
nTempMax = pRoot->m_pleft->m_nMaxRight;

pRoot->m_nMaxLeft = nTempMax + 1;
}

if(pRoot->m_pright != NULL)
{
int nTempMax = 0;

if(pRoot->m_pright->m_nMaxLeft > pRoot->m_pright->m_nMaxRight)
nTempMax = pRoot->m_pright->m_nMaxLeft;
else
nTempMax = pRoot->m_pright->m_nMaxRight;

pRoot->m_nMaxRight = nTempMax + 1;
}

if(pRoot->m_nMaxLeft + pRoot->m_nMaxRight > nMaxLen)
nMaxLen = pRoot->m_nMaxLeft + pRoot->m_nMaxRight;
}

void addBiTreeNode(BiTreeNode *&pCurrent, int value)
{
if(pCurrent == NULL)
{
BiTreeNode *pBiTree = new BiTreeNode();
pBiTree->m_nValue = value;
pBiTree->m_pleft = NULL;
pBiTree->m_pright = NULL;
pCurrent = pBiTree;
}
else
{
if((pCurrent->m_nValue) > value)
addBiTreeNode(pCurrent->m_pleft, value);
else if((pCurrent->m_nValue) < value)
addBiTreeNode(pCurrent->m_pright, value);
}
}