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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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...
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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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Sort Colors II

思路： 才有quicksort的思想，先求出一个midcolor =(1+k)/2。把小于等于mid放左边，大于mid放右边。然后再分解。 实现： 代码如下： public class Solution { /* * @param colors: A list of integer * @param k: An integer * @return: nothing */ public void sortColors2(int[] colors, int k) { // write your code here rainbowSort(colors, 0, colors.length - 1, 1, k); } private void rainbowSort(int[] colors, int left, int right, int fromColor, int toColor) { if(fromColor == toColor) return; if(left >= right) return; int midColor = (fromColor + toColor) / 2; int index = left; for(int i = left; i <= right; i++) { if(colors[i] <= midColor) { swap(colors, i, index++); } } rainbowSort(colors, left, index - 1, fromColor, midColor); rainbowSort(colors, index, right, midColor + 1, toColor); } ...
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