For the given tree, in order traverse is: visit left side root visit right side // 6,8,10,11,12,15,16,17,20,25,27 The successor is the one right next to the target: // target 8 --> succossor is 10 So, given the tree and target node, to find its succe…

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…

The solution for the problem can be divided into three cases: case 1: if the delete node is leaf node, then we can simply remove it case 2: if the delete node is has single side or substree case 3: it has two children, then to keep it as a valid bina…

Problem Link: https://oj.leetcode.com/problems/recover-binary-search-tree/ We know that the inorder traversal of a binary search tree should be a sorted array. Therefore, we can compare each node with its previous node in the inorder to find the two…

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…

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…

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 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…

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…