每日一题-后继者

设计一个算法，找出二叉搜索树中指定节点的“下一个”节点（也即中序后继）。

如果指定节点没有对应的“下一个”节点，则返回null。

示例 1:

输入: root = [2,1,3], p = 1

2
/
1 3

输出: 2
示例 2:

输入: root = [5,3,6,2,4,null,null,1], p = 6

  5
 / \
3   6

/
2 4
/
1

输出: null

java代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
        TreeNode prev = null, curr = root;
        while (!stack.isEmpty() || curr != null) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            if (prev == p) {
                return curr;
            }
            prev = curr;
            curr = curr.right;
        }
        return null;
    }
}