Ugly Number II

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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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515. Find Largest Value in Each Tree Row

Problem: You need to find the largest value in each row of a binary tree. Example: Input: 1 / \ 3 2 / \ \ 5 3 9 Output: [1, 3, 9] Analysis: 这道题是无脑用BFS。居然有人装逼用DFS,不知道怎么想的。 Solution: class Solution { public List< Integer > largestValues ( TreeNode root) { List< Integer > res = new ArrayList<> (); if (root == null) return res; Queue< TreeNode > queue = new LinkedList<> (); queue . offer(root); while ( ! queue . isEmpty()) { int size = queue . size(); int max = Integer . MIN_VALUE; for ( int i = 0 ; i < size; i ++ ) { TreeNode cur = queue . poll(); max = Math . max(max, cur . val); if (cur . left != null) queue . offer(cur . left); if (cur . right != null) queue . offer(...
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27. Remove Element

Problem: Given an array and a value, remove all instances of that value  in-place  and return the new length. Do not allocate extra space for another array, you must do this by  modifying the input array  in-place  with O(1) extra memory. The order of elements can be changed. It doesn't matter what you leave beyond the new length. Example: Given nums = [3,2,2,3], val = 3, Your function should return length = 2, with the first two elements of nums being 2. 5/26/2018 update: 第三遍虽然一开始没有思路，但是稍微想了一下就出来了。其实不用交换，直接赋值。 j++过后又2种情况。 1. nums[j] == nums[i] 2. nums[j] == nums[j - 1] 重复            12223  index j             i 这是第一种情况, j++ 过后是1， nums[i] == nums[j] == 2.            12223  index   j                     i 第二种起情况， j++ 过后是2， nums[j] == nums[j-1], nums[j]在这里是重复的，可以覆盖。 Analysis: ...
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