230. Kth Smallest Element in a BST

Analysis:
Use DFS, it's intuitive.

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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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5. Longest Palindromic Substring

Problem: Given a string  s , find the longest palindromic substring in  s . You may assume that the maximum length of  s  is 1000. Example: Input: "babad" Output: "bab" Note: "aba" is also a valid answer. Example: Input: "cbbd" Output: "bb" 5/22/2018 update: No need to think about after while loop, whether start and end are valid or not. Just update palindrome length within the loop. class Solution { int max = 1 ; int start = 0 ; public String longestPalindrome ( String s) { if (s == null || s . length() == 0 ) return "" ; for ( int i = 0 ; i < s . length() - 1 ; i ++ ) { computePalindrome(s, i, i); computePalindrome(s, i, i + 1 ); } return s . substring(start, start + max); } private void computePalindrome ( String s, int l, int r) { int temp = 0 , len = 0 ; while (l ...
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