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? 题意: 二叉树中序遍历 Solution1: Recursion code class Soluti…
Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what"{1,#,2,3}"means? > read m…
题目链接 题目大意:后序遍历二叉树. 法一:普通递归,只是这里需要传入一个list来存储遍历结果.代码如下(耗时1ms): public List<Integer> postorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); list = dfs(root, list); return list; } public static List<Integer>…
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [,,] \ / Output: [,,] Follow up: Recursive solution is trivial, could you do it iteratively? 题目中要求使用迭代用法,利用栈的“先进后出”特性来实现中序遍历. 解法一:(迭代)将根节点压入栈,当其左子树存在时,一直将其左子树压入栈,…
二叉树遍历(前序.中序.后序.层次.深度优先.广度优先遍历) 描述 解析 递归方案 很简单,先左孩子,输出根,再右孩子. 非递归方案 因为访问左孩子后要访问右孩子,所以需要栈这样的数据结构. 1.指针指向根,根入栈,指针指向左孩子.把左孩子当作子树的根,继续前面的操作. 2.如果某个节点的左孩子不存在,节点出栈,指针指向节点的右孩子.把这个右节点当作根, 继续前面的操作. 代码 /** * Definition for a binary tree node. * public class Tre…
Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively?中序遍历二叉树,递归遍历当然很容易,题目还要求不用递归,下面给出两种方法: 递归: /**…
