*/classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);current=current.right;}// if there is left, get the rightmost ...
1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12vector<int> inorderTraversal(TreeNode *root) {13vector<int>ret;14if(root ==...
65publicstaticvoidinOrder(TreeNode root){66if(root ==null)return;67inOrder(root.left);68visit(root);69inOrder(root.right);70}7172publicstaticvoidinOrder2(TreeNode root){73if(root ==null)return;74Stack<TreeNode> stack =newStack<TreeNode>();75while(!stack.empty() || root !=null){76...
中序Inorder: 先访问左子树,然后访问根节点,最后访问右子树. 后序Postorder:先访问左子树,然后访问右子树,最后访问根节点. classNode:def__init__(self,key):self.left=Noneself.right=Noneself.val=keydefprintInorder(root):ifroot:printInorder(root.left)print(root.val)printInorder(root.right)defprintP...
Preorder inorder postorder 这三种tree traversal一定要记住,属于基本功。自己平时没事就练习一下,有助于打好基础 preorder: root, left, right inorder: left root right postorder: left right root 今天重新做了buildTree 系列,从inorder, postorder中buildTree。 从inorder, preorder 中buildtree。 还有从...
Given preorder and inorder traversal of a tree, construct the binary tree. **Note:**You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: ...
Given a binary tree, return thepreordertraversal of its nodes' values. Example: Input:[1,null,2,3]1 \ 2 / 3Output:[1,2,3] Follow up:Recursive solution is trivial, could you do it iteratively? 给定一个二叉树,返回它的前序遍历。
15、课程:树(下).10、练习—Iterative Postorder Traversal -- -- 10:33 App 15、课程:树(下).7、练习—Iterative Get和Iterative Add 2 -- 12:36 App 15、课程:树(下).13、练习—Construct Binary Tree from Preorder and Inorder Traversal 1 -- 9:07 App 15、课程:树(下).3、练习—Floor and...
上边的两种解法,空间复杂度都是O(n),利用 Morris Traversal 可以使得空间复杂度变为O(1)。 它的主要思想就是利用叶子节点的左右子树是null,所以我们可以利用这个空间去存我们需要的节点,详细的可以参考94 题中序遍历。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Tree...
A.1 B.2 C.3 D.4 暂无答案