求二叉树两结点的最近共同父结点

求二叉树两结点的最近共同父结点，在网上看到了一个挺有意思的解法，原文在http://www.cnblogs.com/remlostime/archive/2012/11/26/2788795.html

一般思路是从子节点深度遍历到根节点，然后比较两个路径的分叉点。他用的递归：

 1 struct Node

 2 {

 3     int val;

 4     Node *left;

 5     Node *right;

 6     Node():left(NULL), right(NULL){}

 7 };

 8 

 9 Node *findFather(Node *root, Node *node1, Node *node2, bool &node1Find, bool &node2Find)

10 {

11     if (root == NULL)

12         return NULL;

13 

14     bool leftNode1Find = false, leftNode2Find = false;

15     Node *leftNode = findFather(root->left, node1, node2, leftNode1Find, leftNode2Find);

16     if (leftNode != NULL)

17         return leftNode;

18 

19     bool rightNode1Find = false, rightNode2Find = false;

20     Node *rightNode = findFather(root->right, node1, node2, rightNode1Find, rightNode2Find);

21     if (rightNode != NULL)

22         return rightNode;

23 

24     node1Find = leftNode1Find || rightNode1Find;

25     node2Find = leftNode2Find || rightNode2Find;

26 

27     if (node1Find && node2Find)

28         return root;

29 

30     if (root == node1)

31         node1Find = true;

32 

33     if (root == node2)

34         node2Find = true;

35 

36     return NULL;

37 }