只需要考虑这个结点是否有右子树

1. 结点有右子树，那么下一个结点就是其右子树的最左结点
2. 结点没有右子树，但是结点是其父结点的左结点，那么其下一个结点就是其父结点
3. 结点没有右子树，又是父结点的右结点，那么就向上遍历，找到一个是父结点左儿子的结点，如果存在，则这个
   父结点为所求结点

后两种情况可以放在一起看。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode father;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode inorderSuccessor(TreeNode p) {

        TreeNode nextNode = null;

        if (p.right != null) {
            TreeNode rightNode = p.right;
            while (rightNode.left != null) {
                rightNode = rightNode.left;
            }
            nextNode = rightNode;
        }
        else if (p.father != null) {
            TreeNode currentNode = p;
            TreeNode fatherNode = p.father;

            while (fatherNode != null && currentNode == fatherNode.right) {
                currentNode = fatherNode;
                fatherNode = currentNode.father;
            }
            nextNode = fatherNode;
        }

        return nextNode;
    }
}