题意 略 分析 1.首先要了解到BST的中序遍历是递增序列 2.我们用一个临时节点tmp储存p的中序遍历的下一个节点,如果p->right不存在,那么tmp就是从root到p的路径中大于p->val的最小数,否则就遍历p的右子树,找到最左边的节点即可 代码 class Solution { public: /* * @param root: The root of the BST. * @param p: You need find the successor node of p. * @re…

Given a binary search tree (See Definition) and a node in it, find the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, returnnull. 分析: 给一个二叉查找树,以及一个节点,求该节点的中序遍历后继,如果没有返回 null. 一棵BST定义为: 节点的左子树中的值要严…

［抄题］: 给一个二叉查找树以及一个节点,求该节点的中序遍历后继,如果没有返回null ［思维问题］: 不知道分合算法和后序节点有什么关系:直接return表达式就行了,它自己会终止的. ［一句话思路］: 比root大时直接扔右边递归,比root小时 考虑是左边递归还是就是root ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 要定义left节点,留着做后续的比较 ［二刷］: ［三刷］: ［四刷］: ［五刷］: ［总结］…

struct node { int val; node *left; node *right; node *parent; node() : val(), left(NULL), right(NULL) { } node(int v) : val(v), left(NULL), right(NULL) { } }; node* insert(int n, node *root) { if (root == NULL) { root = new node(n); return root; } if…

iven a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.…

题目: Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with…

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys…

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. Hide Tags Depth-first Search Linked List       这题是将链表变成二叉树,比较麻烦的遍历过程,因为链表的限制,所以深度搜索的顺序恰巧是链表的顺序,通过设置好递归函数的参数,可以在深度搜索时候便可以遍历了.   TreeNode * he…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 解题思路: 1. 找到数组的中间节点将其置为根节点 2. 左边的即为左子树 3. 右边的即为右子树 4. 递归求解 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * Tr…

Two methods: 1. Traverse 2. Divide & Conquer // Traverse: usually do not have return value public class Solution { public void traverse(TreeNode root) { if (root == null) return; traverse(root.left); traverse(root.right); } } // Divide & Conquer:…

Verify Preorder Sequence in Binary Search Tree \Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree. You may assume each number in the sequence is unique. Follow up: Could you do it using on…

既上篇关于二叉搜索树的文章后,这篇文章介绍一种针对二叉树的新的中序遍历方式,它的特点是不需要递归或者使用栈,而是纯粹使用循环的方式,完成中序遍历. 线索二叉树介绍 首先我们引入“线索二叉树”的概念: "A binary tree is threaded by making all right child pointers that would normally be null point to the inorder successor of the node, and all left chi…

Binary Tree: 0到2个子节点; Binary Search Tree: 所有左边的子节点 < node自身 < 所有右边的子节点: 1. Full类型: 除最下面一层外, 每一层都有两个子节点: 2. Complete类型: 除最下面一层外为Full类型, 但是最下面一层最所有子节点靠左: 3. Balanced类型: 左右两个子树的长度相差小于等于一: /** * Definition of TreeNode: * public class TreeNode { * public…

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? 使用O(n)空间的话可以直接中序遍历来找问题节点. 如果是…

Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order. Example…

Given a binary search tree, print the elements in-order iteratively without using recursion. Note:Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, c…

https://leetcode.com/problems/binary-search-tree-iterator/ Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: nex…

Validate Binary Search Tree Total Accepted: 23828 Total Submissions: 91943My Submissions Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with…

1.题目描述 2.问题分析 首先将二叉查找树使用中序遍历的方式将元素放入一个vector,然后在vector 中截取符合条件的数字. 3.代码 /** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; *…

144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…

题目: Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order. If k…

Validate Binary Search Tree Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node c…

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key.    The right subtree of a node contains only nodes with k…

Given a binary search tree and a new tree node, insert the node into the tree. You should keep the tree still be a valid binary search tree. Example Given binary search tree as follow: 2 / \ 1 4 / 3 after Insert node 6, the tree should be: 2 / \ 1 4…

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?   法I:BST的中序遍历结果是递增序列.把这个递增序列…

题目: Given a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after rem…

二叉树(Binary Tree)是最简单的树形数据结构,然而却十分精妙.其衍生出各种算法,以致于占据了数据结构的半壁江山.STL中大名顶顶的关联容器--集合(set).映射(map)便是使用二叉树实现.由于篇幅有限,此处仅作一般介绍(如果想要完全了解二叉树以及其衍生出的各种算法,恐怕要写8~10篇). 1)二叉树(Binary Tree) 顾名思义,就是一个节点分出两个节点,称其为左右子节点:每个子节点又可以分出两个子节点,这样递归分叉,其形状很像一颗倒着的树.二叉树限制了每个节点最多有两个子节…

原题链接在这里:https://leetcode.com/problems/binary-search-tree-iterator/ Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. N…

题目: Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target. Note: Given target value is a floating point. You may assume k is always valid, that is: k ≤ total nodes. You are guaranteed to have…

Closest Binary Search Tree Value  Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. Note: Given target value is a floating point. You are guaranteed to have only one unique value in the…