Process analysis Stack = 5,  Push 3, Stack = 5, 3.    Pre = 5 Current = 3, Pre = 5, Push 2 to the stack, stack = 5, 3, 2, Pre = 3 Current = 2, Pre = 3, pop 2, PRINT 2, Stack = 5, 3.  Pre = 2 Current = 3, push 4, Stack = 5, 3 , 4.  Pre = 3 Current = 4…
Binary Tree Level Order Traversal Total Accepted: 79463 Total Submissions: 259292 Difficulty: Easy 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,…
Binary Tree Zigzag Level Order Traversal 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,…
/* Author: Jiangong SUN */ Here I will introduce the breadth first traversal of binary tree. The principe is that you traverse the binary tree level by level. This traversal is quite different than depth first traversal. In depth first traversal you…
Binary Tree Level Order Traversal OJ: https://oj.leetcode.com/problems/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…
Title: 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,…
Binary Tree Level Order Traversal 题目链接 题目要求: 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…
1. Unique Binary Search Trees Given n, how many structurally unique BST's (binary search trees) that store values 1...n? For example,Given n = 3, there are a total of 5 unique BST's. 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3 思路:和 Unique Binar…
今天在切leetcode的时候看到一个Morris算法,用来中序遍历二叉树,非递归,O(1)空间.觉得很强大.记录一下. 基本思想是利用了Threaded Binary Tree. 步骤如下: current节点设置为root.如果current不为空,到2,否则返回: 如果current没有左子树,输出current的值,current等于current.right: 如果current有左子树,首先找到current节点的precedent,也就是该节点左子树中最最右边那个节点.然后把最最右…
1.二叉树定义 // Definition for a binary tree node. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; 2.遍历 a.递归先序: //递归先序: 中左右.PS:中序-左中右,后序-左右中,调换cout的位置即可 void NLR(TreeNode* T) { if(T!=NULL…