Minimum Subtree

思路：
DC和traversal 结合。用全局变量记录临时的min sum和sub tree。

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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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3. Longest Substring Without Repeating Characters

Problem: Given a string, find the length of the  longest substring  without repeating characters. Examples: Given  "abcabcbb" , the answer is  "abc" , which the length is 3. Given  "bbbbb" , the answer is  "b" , with the length of 1. Given  "pwwkew" , the answer is  "wke" , with the length of 3. Note that the answer must be a  substring ,  "pwke"  is a  subsequence  and not a substring.
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42. Trapping Rain Water

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Problem: Given  n  non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given  [0,1,0,2,1,0,1,3,2,1,2,1] , return  6 . Analysis: https://leetcode.com/problems/trapping-rain-water/solution/ Solution: class Solution { public int trap ( int [] height) { int left = 0 ; int right = height . length - 1 ; int leftMax = 0 ; int rightMax = 0 ; int res = 0 ; while (left < right) { if (height[left] < height[right]) { leftMax = Math . max(leftMax, height[left]); res += leftMax - height[left]; left ++ ; } else { rightMax = Math . max(rightMax, height[right]); res += rightMax - height[right]; right -- ; } } return res; } }
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