编程之美--3.8

题目描述：求二叉树节点的最大距离，距离是节点之间边的数目

思路：递归判断左子树右子树以及经过当前节点的值的大小

#include <iostream>
#include <queue>
#include <climits>
#include <algorithm>
#include <memory.h>
#include <stdio.h>
#include <ostream>
#include <vector>
#include <list>
#include <cmath>
#include <string>
#include <stdexcept>
#include <stack>
using namespace std;

struct Node
{
    Node* leftChild;
    Node* rightChild;
    int val;
    Node():leftChild(NULL),rightChild(NULL)
    {

    }
};

int getTreeheight(Node *root)
{
    if(root == NULL)
        return 0;
    int left = getTreeheight(root->leftChild)+1;
    int right = getTreeheight(root->rightChild)+1;
    return max(left,right);
}

int fun(Node *root)
{
    if(root == NULL)
        return 0;
    int maxlen = getTreeheight(root->leftChild)+getTreeheight(root->rightChild);
    int tmp = max(fun(root->leftChild),fun(root->rightChild));
    return max(maxlen,tmp);
}

int main()
{

    return 0;
}

 有个更好的解法：时间和空间都优化了

http://www.cnblogs.com/miloyip/archive/2010/02/25/1673114.html