后续遍历-二叉树的最近公共祖先

236. 二叉树的最近公共祖先
要点：
1.自低向上==> 后序遍历

2. 根据左右子节点是否分别是p,q.判断当前节点是否是公共祖先
3. 当前节点是p,q 则返回当前节点。

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

    if (left && right)
        return root;
    else if (left)
        return left;
    else
        return right;
}