【LeetCode 236】Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______

       /              \

    ___5__          ___1__

   /      \        /      \

   6      _2       0       8

         /  \

         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

思路：

　　通过先序遍历，分别找到要查找结点（5 , 4）的路径（3->5 , 3->5->2->4），然后求路径序列中最后一个公共结点（5）即为所求。

C++:

 1 /**

 2  * Definition for a binary tree node.

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 private:

12     vector<TreeNode* > plist;

13     

14 public:

15     void rec(TreeNode *root, TreeNode *tar, vector<TreeNode* >& res)

16     {

17         if(root == tar)

18         {

19             vector<TreeNode* >::iterator it = plist.begin();

20             for(; it != plist.end(); it++)

21                 res.push_back(*it);

22             

23             return ;

24         }

25         

26         if(root->left != 0)

27         {

28             plist.push_back(root->left);

29             rec(root->left, tar, res);

30             plist.pop_back();

31         }

32         

33         if(root->right != 0)

34         {

35             plist.push_back(root->right);

36             rec(root->right, tar, res);

37             plist.pop_back();

38         }

39     }

40 

41     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

42         if(root == 0)

43             return 0;

44         

45         vector<TreeNode* > list1, list2;

46         

47         plist.push_back(root);

48         rec(root, p, list1);

49         

50         plist.clear();

51         plist.push_back(root);

52         rec(root, q, list2);

53         

54         TreeNode  *ret = 0;

55         vector<TreeNode* >::iterator it1 = list1.begin();

56         vector<TreeNode* >::iterator it2 = list2.begin();

57         for(; it1 != list1.end() && it2 != list2.end(); it1++, it2++)

58         {

59             if(*it1 == *it2)

60                 ret = *it1;

61         }

62         

63         return ret;

64     }

65 };