【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 };