LeetCode 285. Inorder Successor in BST

最新推荐文章于 2022-01-16 05:43:31 发布

jmspan

最新推荐文章于 2022-01-16 05:43:31 发布

阅读量1.2k

点赞数

CC 4.0 BY-SA版权

分类专栏：二叉树二叉树遍历二叉搜索树查找搜索比较栈中间状态文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/jmspan/article/details/51156370

二叉树同时被 3 个专栏收录

52 篇文章

订阅专栏

二叉树遍历

22 篇文章

订阅专栏

栈

22 篇文章

订阅专栏

本文介绍三种解决二叉搜索树中序后继问题的方法。第一种为普通二叉树中序遍历；第二种利用BST特性结合栈保存状态；第三种同样利用BST特性，但通过判断节点位置优化时间复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题网址：https://leetcode.com/problems/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.

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

方法一：普通的二叉树中序遍历，没有使用到BST特性，时间复杂度O(n)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode prev, p, succ;
    private void inorder(TreeNode root) {
        if (succ != null) return;
        if (root.left != null) inorder(root.left);
        if (prev == p) {
            succ = root;
            prev = root;
            return;
        }
        prev = root;
        if (root.right != null) inorder(root.right);
    }
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if (root == null) return null;
        this.p = p;
        inorder(root);
        return succ;
    }
}

方法二：利用BST特性，通过栈来保存中间状态。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    Stack<TreeNode> stack = new Stack<>();
    
    private boolean find(TreeNode root, TreeNode p) {
        stack.push(root);
        if (root == p) return true;
        if (root.val < p.val) return find(root.right, p);
        return find(root.left, p);
    }
    
    private TreeNode leftmost(TreeNode root) {
        if (root.left != null) return leftmost(root.left);
        return root;
    }
    
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        find(root, p);
        TreeNode node = stack.pop();
        if (node==p && node.right != null) return leftmost(node.right);
        while (!stack.isEmpty()) {
            TreeNode prev = node;
            node = stack.pop();
            if (prev == node.left) return node;
        }
        return null;
    }
}

方法三：利用BST特性判断节点的可能位置，时间复杂度O(logn)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode prev, p, succ;
    private void find(TreeNode root) {
        if (succ != null) return;
        if (root.val == p.val) {
            prev = root;
            if (root.right != null) find(root.right);
        } else if (root.val < p.val) {
            if (root.right != null) find(root.right);
        } else {
            // root.val > p.val
            if (root.left != null) find(root.left);
            if (prev != null && succ == null) {
                succ = root;
            }
        }
    }
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if (root == null) return null;
        this.p = p;
        find(root);
        return succ;
    }
}

另一种实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean found = false;
    private TreeNode p;
    private TreeNode succ;
    private void find(TreeNode node) {
        if (node == null) return;
        if (succ != null) return;
        if (p.val < node.val) find(node.left);
        if (node == p) found = true;
        else if (found && succ == null) succ = node;
        if (p.val >= node.val) find(node.right);
    }
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        this.p = p;
        find(root);
        return succ;
    }
}