103. 二叉树的锯齿形层次遍历 103. Binary Tree Zigzag Level Order Traversal 题目描述 给定一个二叉树,返回其节点值的锯齿形层次遍历.(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行). LeetCode103. Binary Tree Zigzag Level Order Traversal中等 例如: 给定二叉树 [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回锯齿形层次遍历如…

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…

二叉树(Binary Tree)是最简单的树形数据结构,然而却十分精妙.其衍生出各种算法,以致于占据了数据结构的半壁江山.STL中大名顶顶的关联容器--集合(set).映射(map)便是使用二叉树实现.由于篇幅有限,此处仅作一般介绍(如果想要完全了解二叉树以及其衍生出的各种算法,恐怕要写8~10篇). 1)二叉树(Binary Tree) 顾名思义,就是一个节点分出两个节点,称其为左右子节点:每个子节点又可以分出两个子节点,这样递归分叉,其形状很像一颗倒着的树.二叉树限制了每个节点最多有两个子节…

1.二叉树(Binary Tree) 是n(n>=0)个结点的有限集合,该集合或者为空集(空二叉树),或者由一个根节点和两棵互不相交的,分别称为根节点的左子树和右子树的二叉树组成.  2.特数二叉树 1)斜二叉树 所有的结点都只有左子树的二叉树叫做左斜树 所有的结点都只有右子树的二叉树叫做右斜树 相当于链表,所以线性结构可以理解为是树的一种极其特殊的表现形式 2)满二叉树(完美二叉树) 所有分支结点都存在左子树和右子树. 所有叶子结点都在同一层 3)完全二叉树 对树按照上->下,左->右…

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…

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? 给定一个二叉树,返回它的中序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出…

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,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 retur…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…