Problem link: http://oj.leetcode.com/problems/linked-list-cycle/ We set two pointers: the faster pointer goes two steps each iteration, and the slower one goes one step. If the two pointers meet some time, it follows that there is a loop; otherwise,…

Problem link: http://oj.leetcode.com/problems/linked-list-cycle-ii/ The solution has two step: Detecting the loop using faster/slower pointers. Finding the begining of the loop. After two pointers meet in step 1, keep one pointer and set anther to th…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space?   Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Fol…

Problem Link: http://oj.leetcode.com/problems/interleaving-string/ Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example, Given: s1 = "aabcc", s2 = "dbbca", When s3 = "aadbbcbcac", return t…

Problem link: http://oj.leetcode.com/problems/reverse-words-in-a-string/ Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". LeetCode OJ supports Python now! The s…

Problem Link: http://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/ The problem is asking for flatterning a binary tree to linked list by the pre-order, therefore we could flatten tree from the root. For each node, we link it with its n…

Problem Link: http://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/ We design a auxilar function that convert a linked list to a node with following properties: The node is the mid-node of the linked list. The node's left child i…

Problem link: http://oj.leetcode.com/problems/copy-list-with-random-pointer/ Deepcopy a linked list with random pointer, realy simple problem, solved in three steps using a hash map: Create a new head, duplicating the head (return NULL for case head=…

Problem Link: https://oj.leetcode.com/problems/validate-binary-search-tree/ We inorder-traverse the tree, and for each node we check if current_node.val > prev_node.val. The code is as follows. # Definition for a binary tree node # class TreeNode: #…

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…

Problem Link: https://oj.leetcode.com/problems/same-tree/ The following recursive version is accepted but the iterative one is not accepted... # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left =…

Problem Link: https://oj.leetcode.com/problems/symmetric-tree/ To solve the problem, we can traverse the tree level by level. For each level, we construct an array of values of the length 2^depth, and check if this array is symmetric. The tree is sym…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Traverse the tree level by level using BFS method. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # se…

Problem Link: https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Just BFS from the root and for each level insert a list of values into the result. # Definition for a binary tree node # class TreeNode: # def __init__(self, x):…

Problem Link: https://oj.leetcode.com/problems/maximum-depth-of-binary-tree/ Simply BFS from root and count the number of levels. The code is as follows. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x #…

Problem Link: https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ The basic idea is same to that for Construct Binary Tree from Inorder and Postorder Traversal. We solve it using a recursive function. First, we…

Problem Link: https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ This problem can be easily solved using recursive method. By given the inorder and postorder lists of the tree, i.e. inorder[1..n] and postorde…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/ Use BFS from the tree root to traverse the tree level by level. The python code is as follows. # Definition for a binary tree node # class TreeNode: # def __init__(s…

Problem Link: http://oj.leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ Same idea to Convert Sorted Array to Binary Search Tree, but we use a recursive function to construct the binary search tree. # Definition for a binary tree nod…

Problem Link: http://oj.leetcode.com/problems/balanced-binary-tree/ We use a recursive auxilar function to determine whether a sub-tree is balanced, if the tree is balanced, it also return the depth of the sub-tree. A tree T is balanced if the follow…

Problem Link: http://oj.leetcode.com/problems/minimum-depth-of-binary-tree/ To find the minimum depth, we BFS from the root and record the depth. For each level we add 1 to the depth and return the depth value when we reach a leaf. The python code is…

Problem Link: http://oj.leetcode.com/problems/path-sum-ii/ The basic idea here is same to that of Path Sum. However, since the problem is asking for all possible root-to-leaf paths, so we should use BFS but not DFS. The python code is as follows. # D…

Problem Link: http://oj.leetcode.com/problems/path-sum/ One solution is to BFS the tree from the root, and for each leaf we check if the path sum equals to the given sum value. The code is as follows. # Definition for a binary tree node # class TreeN…

Problem Link: http://oj.leetcode.com/problems/distinct-subsequences/ A classic problem using Dynamic Programming technique. Let m and n be the length of the strings T and S. Let R[i][j] be the count of distinct subsequence of T[0..i] in S[0..j]. Obvi…

Problem Link: http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ OK... Exactly same to Populating Next Right Pointers in Each Node.…

Problem Link: http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/ Just traverse the tree from the root, level by level. Esay enough. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val =…

Problem Link: http://oj.leetcode.com/problems/pascals-triangle-ii/ Let T[i][j] be the j-th element of the i-th row in the triangle, for 0 <= j <= i, i = 0, 1, ... And we have the recursive function for T[i][j] T[i][j] = 1, if j == 0 or j == i T[i][j…

Prolbem Link: http://oj.leetcode.com/problems/pascals-triangle/ Just a nest-for-loop... class Solution: # @return a list of lists of integers def generate(self, numRows): # Initialize the triangle res = [] for i in xrange(numRows): res.append([1] * (…

Problem Link: http://oj.leetcode.com/problems/triangle/ Let R[][] be a 2D array where R[i][j] (j <= i) is the minimum sum of the path from triangle[0][0] to tirangle[i][j]. We initialize R[0][0] = triangle[0][0], and update R[][] from i = 1 to n-1: R…

Problem Link: http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/ Linear Time Solution We try to solve this problem in O(n) time in the help of the algorithm in Best Time to Buy and Sell Stock, which can return the max profit by give…