https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
由于是二叉排序树，所以可以先确定一个搜索函数，方便构造出一个数组用于存放搜寻路径
之后再搜索路径中查找到最后重复的节点即可。
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 11 12 class Solution { 13 public: 14 void getPath(TreeNode * root,TreeNode * p,vector<TreeNode *> & path) 15 { 16 TreeNode * temp=root; 17 while(temp!=NULL) 18 { 19 path.push_back(temp); 20 if(p->val==temp->val) 21 break; 22 else if(p->val>temp->val) 23 temp=temp->right; 24 else 25 temp=temp->left; 26 } 27 } 28 int min(int a,int b) 29 { 30 return a<b?a:b; 31 } 32 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 33 34 vector<TreeNode *> path1; 35 vector<TreeNode *> path2; 36 getPath(root,p,path1); 37 getPath(root,q,path2); 38 int size1=path1.size(); 39 int size2=path2.size(); 40 int i=0; 41 TreeNode * res=root; 42 for(;i<size1&&i<size2;i++) 43 { 44 if(path1.at(i)->val==path2.at(i)->val) 45 { 46 res=path1.at(i); 47 } 48 else 49 break; 50 } 51 return res; 52 53 } 54 };