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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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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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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437. Path Sum III

Problem: You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes). The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000. Example: root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11 Analysis: 4/23/2018 update: 用presum的方法更好啊，递归套递归，把自己套晕了。 This problem is underrated, can category to medium. Solution 1 (Top down recursion) : The path sum of root = pathSumFrom(root) + pathSum(root.left) + pathSum(root.right). When implementing pathSumFrom,   when found root.val == sum, don't stop. Cuz' there are cases when the following path's sum would be 0. ...
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