285. Inorder Successor in BST

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

The successor of a node p is the node with the smallest key greater than p.val.


Example 1:

Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of TreeNode type.

Example 2:

Input: root = [5,3,6,2,4,null,null,1], p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is null.



  1. If the given node has no in-order successor in the tree, return null.
  2. It's guaranteed that the values of the tree are unique.

285. Inorder Successor in BST
struct Solution;
use rustgym_util::*;

trait Inorder {
    fn inorder(&self, p: i32, successor: &mut TreeLink);

impl Inorder for TreeLink {
    fn inorder(&self, p: i32, successor: &mut TreeLink) {
        if let Some(node) = self {
            let node = node.borrow();
            node.left.inorder(p, successor);
            if successor.is_none() && node.val > p {
                *successor = tree!(node.val);
            node.right.inorder(p, successor);

impl Solution {
    fn inorder_successor(root: TreeLink, p: TreeLink) -> TreeLink {
        let p = p.as_ref().unwrap().borrow().val;
        let mut res = None;
        root.inorder(p, &mut res);

fn test() {
    let root = tree!(2, tree!(1), tree!(3));
    let p = tree!(1);
    let res = tree!(2);
    assert_eq!(Solution::inorder_successor(root, p), res);
    let root = tree!(5, tree!(3, tree!(2, tree!(1), None), tree!(4)), tree!(6));
    let p = tree!(6);
    let res = None;
    assert_eq!(Solution::inorder_successor(root, p), res);