2.2 last.right 不为 null,说明之前已经访问过,第二次来到这里,表明当前子树遍历完成,保存 cur 的值,更新 cur = cur.right public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left ...
题目描述英文版描述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: []…
1#Definition for a binary tree node.2#class TreeNode(object):3#def __init__(self, x):4#self.val = x5#self.left = None6#self.right = None78classSolution(object):9definorderTraversal(self, root):10"""11:type root: TreeNode12:rtype: List[int]13"""14ifnotroot: # 空节点直接返...
题目:Binary Tree Inorder Traversal 二叉树的中序遍历,和前序、中序一样的处理方式,代码见下: 1structTreeNode {2intval;3TreeNode*left;4TreeNode*right;5TreeNode(intx): val(x), left(NULL),right(NULL) {}6};78vector<int> preorderTraversal(TreeNode *root)//非递归的中序遍历(用栈实现)9{10...
【题目描述】 Given a binary tree, return the inorder traversal of its nodes' values. 给出一棵二叉树,...
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....
The pre-order and post-order traversal of a Binary Tree generates the same output. The tree can have maximum . A、Three nodes B、Two nodes C、One node D、Any number of nodes 暂无答案
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
Today’s practice algorithm question is to invert a binary tree. This is a very good problem to start learning with tree data structure; specifically, binary tree.
tree->left->right = new Node(4); tree->left->right->left = new Node(6); tree->right->right = new Node(5); // Print inorder traversal of the input tree cout <<"Inorder of the given tree: "; inorder(tree); Node* mirrorTree = NULL; ...