Binary Search Notes

用l + 1 < r再二次检查的模板。二次检查要遵循mid check。

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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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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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