*/// 由于主函数的形式已经符合分治法需要的形式(具有合适的返回值),直接使用主函数做为递归函数vector<int>preorderTraversal(TreeNode * root){//递归三要素之定义// write your code herevector<int> result;if(root == nullptr) {returnresult;// 递归三要素之出口
解法三 Morris Traversal 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。 public List<Integer> preorderTraversal(TreeNode root) { Li...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> preorderTraversal(TreeNode*root) { vector<int>result; stack<TreeNode *...
Can you solve this real interview question? Binary Tree Preorder Traversal - Given the root of a binary tree, return the preorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: [https://assets.l
Given a binary tree, return thepreordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[1,2,3]. Note:Recursive solution is trivial, could you do it iteratively?
binary-tree-preorder-traversal 题意:前序遍历二叉树 前序遍历 根->左子树->右子树 先递归解法: classSolution {public: vector<int> preorderTraversal(TreeNode *root) { vector<int>res; position(root, res);returnres; }voidposition(TreeNode *root, vector<int> &res){if(root){...
Morris traversal: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<...
[TreeNode]()varcurr=rootwhile(stack.nonEmpty||curr!=null){while(curr!=null){stack.push(curr)rs.append(curr.value)curr=curr.left}valnode=stack.pop()curr=node.right}rs.toList}defpreorderTraversalV4(root:TreeNode):List[Int]={if(root==null)returnNilvalrs=ListBuffer.empty[Int]valstack=Stack...
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...
I am trying to get the following output Preorder: 50 5 20 15 27 42 35 25 55 70 60 80 95 90 75 Inorder: 5 15 20 25 27 35 42 50 55 60 70 75 80 90 95 Postorder: 5 20 15 27 42 35 25 55 70 60 80 95 90 75 50 Instead I am getting this Preorder: 1361612853125975312...