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
     */
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
        // write code here
        return findh(root, o1, o2)->val;
    }

    TreeNode* findh(TreeNode* root, int o1, int o2)
    {
        if(root == NULL) return NULL;
        if (root->val == o1 || root->val==o2)
            return root;

        TreeNode* ifLeftHave = findh(root->left, o1, o2);
        TreeNode* ifRightHave = findh(root->right, o1, o2);
        if(ifRightHave !=NULL && ifLeftHave != NULL) 
            return root;
        else if (ifLeftHave != NULL)
            return ifLeftHave;
        else
            return ifRightHave;
    }
};