285. Inorder Successor in BST

本文介绍了一种寻找二叉搜索树中给定节点的中序后继的方法。通过遍历树并利用二叉搜索树的特性,可以在O(h)的时间复杂度内找到目标节点的后继节点,其中h为树的高度。
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.

做出来了 但是需要遍历整棵树 

下面的基于二分查找 针对后继节点出现的情况进行分类

public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
    TreeNode succ = null;
	while (root != null) {
        if (p.val < root.val) {
            succ = root;
            root = root.left;
         }
        else
            root = root.right;
    }
  return succ;
}
也可以从二叉树的中序遍历是有序的来考虑 就是找所有大于p的最小节点

#include <iostream>#include <iostream>using namespace std;struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}};class BST {public:TreeNode* root;BST() : root(nullptr) {}void insert(int val) {TreeNode* newNode = new TreeNode(val);if (!root) {root = newNode;return;}TreeNode* curr = root;while (true) {if (val < curr->val) {if (curr->left)curr = curr->left;else {curr->left = newNode;break;}}else if (val > curr->val) {if (curr->right)curr = curr->right;else {curr->right = newNode;break;}}else {delete newNode; return;}}}TreeNode* findMin(TreeNode* node) {while (node->left) node = node->left;return node;}TreeNode* remove(TreeNode* node, int val) {if (!node) return nullptr;if (val < node->val) {node->left = remove(node->left, val);}else if (val > node->val) {node->right = remove(node->right, val);}else {if (!node->left) {TreeNode* temp = node->right;delete node;return temp;}else if (!node->right) {TreeNode* temp = node->left;delete node;return temp;}else {TreeNode* successor = findMin(node->right);node->val = successor->val;node->right = remove(node->right, successor->val);}}return node;}void inorder(TreeNode* node) {if (node) {inorder(node->left);cout << node->val << " ";inorder(node->right);}}void remove(int val) {root = remove(root, val);}void printInorder() {inorder(root);cout << endl;}};int main() {BST bst;int arr[] = { 4, 9, 0, 1, 8, 6, 3, 5, 2, 7 };int n = sizeof(arr) / sizeof(arr[0]);for (int i = 0; i < n; ++i) {bst.insert(arr[i]);}cout << "初始中序遍历: ";bst.printInorder();// 删除关键字为4的节点bst.remove(4);cout << "删除4后中序遍历: ";bst.printInorder();// 删除关键字为5的节点bst.remove(5);cout << "删除5后中序遍历: ";bst.printInorder();return 0;} 对此代码进行详细解释
最新发布
11-02
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