【转】求二叉树两结点的最近共同父结点

原文链接： http://www.cnblogs.com/resonantPineapple/p/4496942.html

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 }

 

转载于:https://www.cnblogs.com/resonantPineapple/p/4496942.html