* @return: Preorder in ArrayList which contains node values. */// 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) {returnresu...
*/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.right;}else{// has left, then find the rightmost of left su...
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] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree ...
题目: Given a binary tree, return thepreordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[1,2,3]. Note:Recursive solution is trivial, could you do it iteratively? 说明: 1)递归和非递归实现,其中非递归有两种方法 2)复杂度,时间O(n),空...
I am implementing Preorder Traversal of Binary Tree (without recursion). The following code runs into an infinite loop. I cannot understand what's happeningvoid Tree::n_preorder() { Node* temp; stack s; cout<<"\nPreorder: "; while(...
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 2 / 3 Output: [1,2,3] Follow up: Recursive solution is trivial, could you do it iteratively? 翻译 给出一棵二叉树,返回其节点值的前序遍历。
题目:Given a binary tree, return the preorder traversal of its nodes‘ values.For example:Given binary tree {1,#,2,3},1 2 / 3 return [1,2,3].Note: Recursive solution is trivial, could you do it iteratively?说明:1)递归和非递归实现,其中非递归有两种方法2...
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?
[LeetCode]Binary Tree Preorder Traversal 题目描述: Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 题目大意: 给定一棵二叉树,返回节点值的先序遍历结果(尽量使用非递归算法)。
详见:https://leetcode.com/problems/binary-tree-preorder-traversal/description/ Java实现: 递归实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } ...