236. 二叉树的最近公共祖先/C++

递归，考虑4种情况。

1. pq不在左右子树，返回NULL
2. pq同时在左子树，返回left
3. pq同时在右子树，返回right
4. pq分别在左右子树，返回root，即为所求

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root || root==p || root==q) return root;
        TreeNode* left=lowestCommonAncestor(root->left,p,q);
        TreeNode* right=lowestCommonAncestor(root->right,p,q);
        
        if(!left && !right) return NULL;
        else if(left && !right) return left;
        else if(!left && right) return right;
        return root;
    }
};