题解 | #在二叉树中找到两个节点的最近公共祖先#

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param o1 int整型 
     * @param o2 int整型 
     * @return int整型
     */
    vector<int> o1_v;
    vector<int> o2_v;
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
        // write code here
        o1_v.push_back(-1);
        o2_v.push_back(-1);
        dfs(root, o1, o1_v);
        dfs(root, o2, o2_v);
        int res=0;
        for(int i=0;i<o1_v.size();i++){
            if(o1_v[i]==o2_v[i]) res=o1_v[i];
            else break;
        }
        return res;
    }
    void dfs(TreeNode* root, int val, vector<int>& v){
        if(v.back()==val) return;
        if(root==nullptr) return;
        v.push_back(root->val);
        dfs(root->left, val, v);
        dfs(root->right, val, v);
        if(v.back()!=val)
            v.pop_back();
    }
};