classSolution {public: vector<int> preorderTraversal(TreeNode*root) {if(!root)return{}; vector<int>res; stack<TreeNode*>s{{root}};while(!s.empty()) { TreeNode*t =s.top(); s.pop(); res.push_back(t->val);if(t->ri
classSolution {public: vector<int> preorderTraversal(TreeNode*root) {if(!root)return{}; vector<int>res; stack<TreeNode*>s{{root}};while(!s.empty()) { TreeNode*t =s.top(); s.pop(); res.push_back(t->val);if(t->right) s.push(t->right);if(t->left) s.push(t->left); ...
12 vector<int> preorderTraversal(TreeNode *root) { 13 vector<int> v; 14 if(root==NULL) 15 return v; 16 stack<TreeNode *> tree; 17 tree.push(root); 18 while(!tree.empty()){ 19 TreeNode *p=tree.top(); 20 v.push_back(p->val); 21 tree.pop(); 22 if(p->right!=NULL) 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? 2.解决方案1 classS...
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...
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<...
Given a binary tree, return the preorder traversal 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? Language:Java /** * Definition for a binary tree node. ...
Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the left subtree, and (iii) Traverse the right subtree. Therefore, the Preorder traversal of the above tree will outputs: ...
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
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