BFS Notes


  1. 0 - n-1 graph的表示方法用二维数组或者list[]
  2. Topological sort 关键是要找indegree和neighbour mapping
  3. Topological sort如果nodes是动态的最好是用map而不是数组。先初始化graph和indegree更好写
  4. 与memory相关比如number of island，在加进queue的时候就需要标记，不然会超时。 
  5. 凡是能不用BFS做的都比BFS做简单。比如能用union find做的。
  6. 在矩阵里面，memorize直接可以in place做，不需要额外空间。
  7. 坐标可以用int[2] 来表示

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776. Split BST

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Problem: Given a Binary Search Tree (BST) with root node  root , and a target value  V , split the tree into two subtrees where one subtree has nodes that are all smaller or equal to the target value, while the other subtree has all nodes that are greater than the target value.  It's not necessarily the case that the tree contains a node with value  V . Additionally, most of the structure of the original tree should remain.  Formally, for any child C with parent P in the original tree, if they are both in the same subtree after the split, then node C should still have the parent P. You should output the root TreeNode of both subtrees after splitting, in any order. Example 1: Input: root = [4,2,6,1,3,5,7], V = 2 Output: [[2,1],[4,3,6,null,null,5,7]] Explanation: Note that root, output[0], and output[1] are TreeNode objects, not arrays. The given tree [4,2,6,1,3,5,7] is represented by the following diagram: 4 / \ ...
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426. Convert Binary Search Tree to Sored Doubly Linked List

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Problem: Convert a BST to a sorted circular doubly-linked list in-place. Think of the left and right pointers as synonymous to the previous and next pointers in a doubly-linked list. Let's take the following BST as an example, it may help you understand the problem better: We want to transform this BST into a circular doubly linked list. Each node in a doubly linked list has a predecessor and successor. For a circular doubly linked list, the predecessor of the first element is the last element, and the successor of the last element is the first element. The figure below shows the circular doubly linked list for the BST above. The "head" symbol means the node it points to is the smallest element of the linked list. Specifically, we want to do the transformation in place. After the transformation, the left pointer of the tree node should point to its predecessor, and the right pointer should point to its successor. We should return the pointer to the first...
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