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] Outpu
题目描述英文版描述Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: Input: root = [3,9,20,null,null,1…
* @param root: A Tree * @return: Inorder in ArrayList which contains node values. */vector<int>inorderTraversal(TreeNode * root){// write your code herevector<int> result; traverse(root, result);returnresult; }// 遍历法的递归函数,没有返回值,函数参数result贯穿整个递归过程用来记录遍历的结...
Given a binary tree, return thelevel ordertraversal 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], [15,7] ] ++++++++++++...
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 ...
Using these new definitions, the leaf nodes in binary tree (a) are nodes 6 and 8; the internal nodes are nodes 1, 2, 3, 4, 5, and 7.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 ...
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 ...
描述: 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}, return its level order trav...
* TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>preorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){if(current.left==null){result.add(current.val);current=current...
public List<Integer> inorderTraversal(TreeNode root) { List<Integer> ans = new ArrayList<>(); getAns(root, ans); return ans; } private void getAns(TreeNode node, List<Integer> ans) { if (node == null) { return; } getAns(node.left, ans); ans.add(node.val); getAns(node.right...