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;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param o1 int整型 
     * @param o2 int整型 
     * @return int整型
     */
    //记录是否找到o的路径
    bool flag = false;
    void dfs(TreeNode* root, vector<int>& path, int o)
    {
        if(false || root == nullptr)
        {
            return;
        }
        path.push_back(root->val);

        if(root->val == o)
        {
            flag = true;
            return;
        }

        dfs(root->left, path, o);
        dfs(root->right, path, o);

        if(flag)
        {
            return;
        }
        path.pop_back();
    }
    
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) 
    {
        vector<int> path1, path2;

        dfs(root, path1, o1);

        flag = false;

        dfs(root, path2, o2);

        int res;
        for(int i = 0; i < path1.size() && i < path2.size(); i++)
        {
            if(path1[i] == path2[i])
            {
                res = path1[i];
            }
            else 
            {
                break;
            }
        }
        return res;
    }
};