题解 | #在二叉树中找到两个节点的最近公共祖先#

没在状态，但还是需要保持一下手感

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

class Solution {
  public:
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
      int res;
      std::vector<int> o1_path, o2_path;
      
      dfs(root, o1_path, o1);
      flag = false;
      dfs(root, o2_path, o2);
      
      for (int i = 0; i < o1_path.size() && i < o2_path.size(); ++i) {
        if (o1_path[i] == o2_path[i]) {
          res = o1_path[i];
        }
      }
      
      return res;
    }
  private:
    void dfs(TreeNode *root, std::vector<int> &path, int target) {
      if (flag || root == nullptr) {
        return ;
      }
      
      path.push_back(root->val);
      
      if (root->val == target) {
        flag = true;
        return ;
      }
      
      dfs(root->left, path, target);
      dfs(root->right, path, target);
      
      if (flag) {
        return ;
      }
      
      path.pop_back();
    }
  private:
    bool flag = false;
};