［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］…

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…

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…

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…

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than or equal to the nod…

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…

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…

Return the root node of a binary search tree that matches the given preorder traversal. (Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has…

二叉搜索树是常用的概念,它的定义如下: 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 greater than the node's key. Both the left and right subtrees must also be binary search…