Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0. The tilt of the …

给定一个二叉树,计算整个树的坡度. 一个树的节点的坡度定义即为,该节点左子树的结点之和和右子树结点之和的差的绝对值.空结点的的坡度是0. 整个树的坡度就是其所有节点的坡度之和. 示例: 输入: 1 / \ 2   3 输出: 1 解释: 结点的坡度 2 : 0 结点的坡度 3 : 0 结点的坡度 1 : |2-3| = 1 树的坡度 : 0 + 0 + 1 = 1 注意: 任何子树的结点的和不会超过32位整数的范围. 坡度的值不会超过32位整数的范围. class Solution { publ…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一…

We are given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1. Return the same tree where every subtree (of the given tree) not containing a 1 has been removed. (Recall that the subtree of a node X is X,…

原题链接在这里:https://leetcode.com/problems/binary-tree-tilt/description/ 题目: Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum…

563. 二叉树的坡度 563. Binary Tree Tilt 题目描述 给定一个二叉树,计算整个树的坡度. 一个树的节点的坡度定义即为,该节点左子树的结点之和和右子树结点之和的差的绝对值.空结点的的坡度是 0. 整个树的坡度就是其所有节点的坡度之和. 每日一算法2019/6/10Day 38LeetCode563. Binary Tree Tilt 示例: 输入: 1 / \ 2 3 输出: 1 解释: 结点的坡度 2 : 0 结点的坡度 3 : 0 结点的坡度 1 : |2-3| = 1…

LeetCode:Binary Tree Level Order Traversal 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 ord…

LeetCode: Binary Tree Traversal 题目:树的先序和后序. 后序地址:https://oj.leetcode.com/problems/binary-tree-postorder-traversal/ 先序地址:https://oj.leetcode.com/problems/binary-tree-preorder-traversal/ 后序算法:利用栈的非递归算法.初始时,先从根节点一直往左走到底,并把相应的元素进栈:在循环里每次都取出栈顶元素,如果该栈顶元素的右…

遍历二叉树   traversing binary tree 线索二叉树 threaded binary tree 线索链表 线索化 1. 二叉树3个基本单元组成:根节点.左子树.右子树 以L.D.R分别表示遍历左子树.访问根节点.遍历右子树 可能的情况6种 排列A3 2 LDR LRD DLR DRL RLD RDL 若限定先左后右 LDR LRD  中根序遍历  后根序遍历 DLR  先根序遍历 先/中/后 序遍历…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Example Given 4 points: (1,2), (3,6), (0,0), (1,3). The maximum number is 3. LeeCode上的原题,可参见我之前的博客Invert Binary Tree 翻转二叉树. 解法一: // Recursion class So…