285. Inorder Successor in BST - Medium

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

Note: If the given node has no in-order successor in the tree, return null.

Example 1:

Input: root = [2,1,3], p = 1

  2
 / \
1   3

Output: 2

Example 2:

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

      5
     / \
    3   6
   / \
  2   4
 /   
1

Output: null

 

M1: recursive

time: O(n), space: O(height)

/**
 * 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) {
        if(root == null) {
            return null;
        }
        if(p.val >= root.val) {
            return inorderSuccessor(root.right, p);
        } else {
            TreeNode left = inorderSuccessor(root.left, p);
            if(left != null) {
                return left;
            } else {
                return root;
            }
        }
    }
}

 

M2: iterative

分两种情况考虑：p有无右子节点

如果p有右子节点，返回右子树的最左子节点；如果没有，从root开始按inorder遍历找successor

time: O(n), space: O(1)

/**
 * 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) {
        if(root == null) {
            return null;
        }
        
        if(p.right != null) {
            TreeNode cur = p.right;
            while(cur.left != null) {
                cur = cur.left;
            }
            return cur;
        }
        
        else {
            TreeNode s = root, t = null;
            while(s.val != p.val) {
                if(p.val <= s.val) {
                    t = s;
                    s = s.left;
                } else {
                    s = s.right;
                }
            }
            return t;
        }
    }
}