北叶青藤

2016年8月7日

摘要： Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of th 阅读全文

posted @ 2016-08-07 05:21 北叶青藤阅读(279) 评论(0) 推荐(0) 编辑

285. Inorder Successor in Binary Search Tree

摘要： Given a binary search tree and a node in it, find the in-order successor of that node in the BST. 1 public class Solution { 2 3 public TreeNode inorde 阅读全文

posted @ 2016-08-07 00:05 北叶青藤阅读(200) 评论(0) 推荐(0) 编辑

2016年8月6日

304. Range Sum Query 2D - Mutable & Immutable

摘要： Range Sum Query 2D - Mutable Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) 阅读全文

posted @ 2016-08-06 23:13 北叶青藤阅读(217) 评论(0) 推荐(0) 编辑

161. One Edit Distance

摘要： Given two strings S and T, determine if they are both one edit distance apart. 分析：https://segmentfault.com/a/1190000003906621 虽然我们可以用Edit Distance的解法，阅读全文

posted @ 2016-08-06 12:42 北叶青藤阅读(273) 评论(0) 推荐(0) 编辑

277. Find celebrity

摘要： Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity. The definition of a celebrity is tha 阅读全文

posted @ 2016-08-06 11:19 北叶青藤阅读(215) 评论(0) 推荐(0) 编辑

Peeking Iterator

摘要： Given an Iterator class interface with methods: next() and hasNext(), design and implement a PeekingIterator that support the peek()operation -- it es 阅读全文

posted @ 2016-08-06 08:26 北叶青藤阅读(236) 评论(0) 推荐(0) 编辑

Zigzag Iterator

摘要： Given two 1d vectors, implement an iterator to return their elements alternately. For example, given two 1d vectors: By calling next repeatedly until 阅读全文

posted @ 2016-08-06 06:14 北叶青藤阅读(247) 评论(0) 推荐(0) 编辑

Compare Version Numbers

摘要： Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assum 阅读全文

posted @ 2016-08-06 05:51 北叶青藤阅读(200) 评论(0) 推荐(0) 编辑

Recover Binary Search Tree

摘要： 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 阅读全文

posted @ 2016-08-06 05:26 北叶青藤阅读(208) 评论(0) 推荐(0) 编辑

Binary Tree Non-recursive Traversal

摘要： Preorder: 因为是preorder traversal, 我们需要先print root，然后左节点，最后右节点，而且root左边子树一定比右边子树先print出来，所以，我们可以先把当前root的右节点压栈，然后把root的左节点压栈，这样每次从栈里取的时候，可以保证左边节点的root先取阅读全文

posted @ 2016-08-06 02:47 北叶青藤阅读(331) 评论(0) 推荐(0) 编辑

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