235. Lowest Common Ancestor of a

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

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    
    bool dfs(TreeNode* root, TreeNode* t, vector<TreeNode*>& v){
        if(NULL == root) return false;
        if(root == t){
            v.push_back(root);
            return true;
        }
        if(dfs(root->left, t, v) || dfs(root->right, t, v)){
            v.push_back(root);
            return true;
        }else return false;
    }
    
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        vector<TreeNode*> v1 = vector<TreeNode*>();
        vector<TreeNode*> v2 = vector<TreeNode*>();
        dfs(root, p, v1);
        dfs(root, q, v2);
        int i = v1.size() - 1;
        int j = v2.size() - 1;
        for(;i>=0 && j>=0 && v1[i] == v2[j];i--, j--);
        return v1[++i];
    }
};