root.left.right =newTreeNode(3); root.right =newTreeNode(6); root.right.left =newTreeNode(5); root.right.right =newTreeNode(7); root.right.right.right =newTreeNode(8); System.out.println(Main21.inorderTraversal(root)); } publicstaticclassTreeNode { intval; TreeNode left; TreeNode ...
* }*/publicclassSolution {/***@paramroot: The root of binary tree. *@return: Inorder in ArrayList which contains node values.*/publicArrayList<Integer>inorderTraversal(TreeNode root) {//write your code hereArrayList<Integer> rst =newArrayList<Integer>();if(root ==null) {returnrst; } tr...
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> inorderTraversal(TreeNode *root) { 13 vector<int> res; 14 if(root==NULL) return res; 15 stack<TreeNode*> s; ...
94. Binary Tree Inorder Traversal 递归的代码是以前数据结构书上常见的: 非递归用stack模拟中序遍历,要理解啊,不能死记硬背。注意while条件和while里面的if。 MAR 25TH 这题今天做98. Validate Binary Search Tree 的时候又来复习了一遍,又忘的差不多了。。记得当时还在考虑为什么while里面要用||而不是&&...
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], 代码语言:javascript 代码运行次数:0 运行 AI代码解释 ...
Unfortunately, the .NET Framework does not contain a binary tree class, so in order to better understand binary trees, let's take a moment to create our own binary tree class. The First Step: Creating a Base Node Class The first step in designing our binary tree class is to create a ...
Unfortunately, the .NET Framework does not contain a binary tree class, so in order to better understand binary trees, let's take a moment to create our own binary tree class. The First Step: Creating a Base Node Class The first step in designing our binary tree class is to create a ...
Integer>result=newArrayList<Integer>();if(root==null)returnresult;inorderTraversal(root,result);returnresult;}privatevoidinorderTraversal(TreeNode root,ArrayList<Integer>result){if(root.left!=null)inorderTraversal(root.left,result);result.add(root.val);if(root.right!=null)inorderTraversal(root....
publicList<Integer>inorderTraversal(TreeNoderoot){List<Integer>ans=newArrayList<>();getAns(root,ans);returnans;}privatevoidgetAns(TreeNodenode,List<Integer>ans){if(node==null){return;}getAns(node.left,ans);ans.add(node.val);getAns(node.right,ans);} 时间复杂度:O(n),遍历每个节点。 空间...
Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is...