[LeetCode] 285. Inorder Successor in BST 二叉搜索树中的中序后继节点

最新推荐文章于 2025-09-10 15:24:29 发布

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最新推荐文章于 2025-09-10 15:24:29 发布

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文章标签：数据结构与算法 java python

原文链接：http://www.cnblogs.com/lightwindy/p/8661613.html

本文介绍了解决二叉搜索树中寻找指定节点的中序后继节点的问题，提供了多种解法，包括使用中序遍历的方法和利用二叉搜索树的特性进行优化的解决方案。

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.

给一个二叉搜索树和它的一个节点，找出它的中序后继节点，如果没有返回null。

解法1: 用中序遍历二叉搜索树，当找到root.val = p.val的时候，返回下一个节点。T: O(n)， S: O(n)

解法2: 利用BST的性质，不用遍历全部元素。比较root.val和p.val，然后在左子树或者右子树中查找。T: O(h)， S: O(1)

如果节点p有右子树，那么p的后继节点为右子树的的最左值（即最后一个没有左子树的节点）；

如果节点p没有右子树，就在二叉搜索树中搜索节点p，并在从上往下的查找过程中依次更新记录比p的val值大的节点。

Java:

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

Java:

public class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if(root == null || p == null) {
            return null;
        }
        TreeNode successor = null;
        while(root != null) {
            if(p.val < root.val) {
                successor = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }
        
        return successor;
    }
}

Java:

public class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        TreeNode node = root, successor = null;
        while (node != null) {
            if (node.val > p.val) {
                successor = node;
                node = node.left;
            } else {
                node = node.right;
            }
        }
        return successor;
    }
}

Python:

class Solution(object):
    def inorderSuccessor(self, root, p):
        """
        :type root: TreeNode
        :type p: TreeNode
        :rtype: TreeNode
        """
        # If it has right subtree.
        if p and p.right:
            p = p.right
            while p.left:
                p = p.left
            return p

        # Search from root.
        successor = None
        while root and root != p:
            if root.val > p.val:
                successor = root
                root = root.left
            else:
                root = root.right

        return successor

C++:

class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        TreeNode *res = NULL;
        while (root) {
            if (root->val > p->val) {
                res = root;
                root = root->left;
            } else root = root->right;
        }
        return res;
    }
};

C++: Recursion

class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        if (!root) return NULL;
        if (root->val <= p->val) {
            return inorderSuccessor(root->right, p);
        } else {
            TreeNode *left = inorderSuccessor(root->left, p);
            return left ? left : root;
        }
    }
};

类似题目：

[LeetCode] 94. Binary Tree Inorder Traversal 二叉树的中序遍历

[LeetCode] 173. Binary Search Tree Iterator 二叉搜索树迭代器