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

 1 #include <iostream>
 2 #include <queue>
 3 #include <climits>
 4 #include <algorithm>
 5 #include <memory.h>
 6 #include <stdio.h>
 7 #include <ostream>
 8 #include <vector>
 9 #include <list>
10 #include <cmath>
11 #include <string>
12 #include <stdexcept>
13 #include <stack>
14 using namespace std;
15
16 struct Node
17 {
18     Node* leftChild;
19     Node* rightChild;
20     int val;
21     Node():leftChild(NULL),rightChild(NULL)
22     {
23
24     }
25 };
26
27 int getTreeheight(Node *root)
28 {
29     if(root == NULL)
30         return 0;
31     int left = getTreeheight(root->leftChild)+1;
32     int right = getTreeheight(root->rightChild)+1;
33     return max(left,right);
34 }
35
36 int fun(Node *root)
37 {
38     if(root == NULL)
39         return 0;
40     int maxlen = getTreeheight(root->leftChild)+getTreeheight(root->rightChild);
41     int tmp = max(fun(root->leftChild),fun(root->rightChild));
42     return max(maxlen,tmp);
43 }
44
45 int main()
46 {
47
48     return 0;
49 }

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

编程之美--3.8