Lowest Common Ancestor of a Binary Tree

做到这个题才发现之前做的关于二叉检索树的写复杂了，其实可以直接根据二叉检索树的特点进行判断（从树根开始，某一节点的值大于待搜的两个节点则在左边找，小于待搜的两个节点则在右边找，否则返回该节点即可）。这道题倒是必须用DFS来解决。

class Solution {public:    //DFS代码    void findNode(TreeNode* root, TreeNode* toFind, vector<TreeNode*> &curPath, bool& find){        if(root == toFind){            curPath.push_back(root);            find = true;            return;        }        if(root == nullptr)            return;        curPath.push_back(root);        if(root -> left != nullptr){            findNode(root -> left, toFind, curPath, find);            if(find == true)                return;            curPath.pop_back();        }        if(root -> right != nullptr){            findNode(root -> right, toFind, curPath, find);            if(find == true)                return;            curPath.pop_back();        }    }    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {//路径放在两个vector中        vector<TreeNode*> pPath, qPath;        TreeNode* result;        bool pFind = false, qFind = false;//查找两个路径        findNode(root, p, pPath, pFind);        findNode(root, q, qPath, qFind);                if(qFind == false || pFind == false)            return nullptr;//从得到的路径查找公共祖先        for(int index = 0; index < pPath.size() && index < qPath.size() && qPath[index] == pPath[index]; index++){            result = pPath[index];        }                return result;    }};