To do


  1. Dynamic progarmming
  2. design
  3. binary tree
  4. linked list
  5. 括号问题
  6. DFS

评论

发表评论

此博客中的热门博文

663. Equal Tree Partition

Problem: Given a binary tree with  n  nodes, your task is to check if it's possible to partition the tree to two trees which have the equal sum of values after removing  exactly  one edge on the original tree. Example 1: Input: 5 / \ 10 10 / \ 2 3 Output: True Explanation: 5 / 10 Sum: 15 10 / \ 2 3 Sum: 15 Example 2: Input: 1 / \ 2 10 / \ 2 20 Output: False Explanation: You can't split the tree into two trees with equal sum after removing exactly one edge on the tree. Note: The range of tree node value is in the range of [-100000, 100000]. 1 <= n <= 10000 Analysis: Get all sums, check whether the collections of sums contains totalSum/2. Do not include total sum to set of sums, if the total sum is 0, it will qualify.  Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 class Solution { public boolean checkEqualTree(TreeNode ro...
阅读全文

776. Split BST

图片
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 / \ ...
阅读全文

426. Convert Binary Search Tree to Sored Doubly Linked List

图片
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...
阅读全文