Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3 Return 6. 对于树,我们可以找到其左右子树中终止于根节点的最大路径值,称为最大半路径值,如果都是正数,则可以把它们以及根节点值相加,如果其中有负数,则舍弃那一边的值(即置为零).如此可以
Given an n-ary tree, return the preorder traversal of its nodes' values. For example, given a 3-ary tree: Return its preorder traversal as: [1,3,5,6,2,4]. Note: Recursive solution is trivial, could you do it iteratively? -----------------------------
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? ----------------------------------------------------
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ] 对每
庞果网编程英雄会上做的一道题:二分查找(非递归),和大家分享一下: public class BinarySearchClass { public static int binary_search(int[] array, int value) { int beginIndex = 0;// 低位下标 int endIndex = array.length - 1;// 高位下标 int midIndex = -1; while (beginIndex <= endIndex) { midInd