题解 | #在二叉树中找到两个节点的最近公共祖先# C++ 解法，后序从根节点回溯

题解 | #在二叉树中找到两个节点的最近公共祖先#
C++ 解法，后序从根节点回溯

/**
 * 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
     */
    TreeNode *dfs(TreeNode *root, int p, int q) {
        if (root == nullptr || root->val == p || root->val == q) return root;
        TreeNode *left = dfs(root->left, p, q);
        TreeNode *right = dfs(root->right, p, q);
        if (left && right) return root;
        if (left) return left;
        if (right) return right;
        return nullptr;
    }
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
        TreeNode *ans = dfs(root, o1, o2);
        return ans->val;
    }
};