#Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = None"""利用堆栈先进后出的特点"""classSolution(object):defpreorderTraversal(self, root):""":type root: TreeNode :rtype: List[int]"""result=[] slack=...
TreeNode(intx) { val = x; } } 第一种方法:算法实现类 importjava.util.LinkedList;importjava.util.List;publicclassSolution{privateList<Integer> result;publicList<Integer>preorderTraversal(TreeNode root){ result =newLinkedList<>(); preOrder(root);returnresult; }privatevoidpreOrder(TreeNode root)...
public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder.length == 0 || inorder.length == 0) return null; return helper(0,inorder.length - 1,preorder,inorder); } private TreeNode helper(int inStart, int inEnd, int[] preorder, int[] inorder){ // Base情况 if...
The function returns the root of the constructed binary tree. 4. Time & Space Complexity Analysis: 4.1 Time Complexity: 4.1.1 Lists as parameters In each recursive call, the index() function is used to find the index of the root value in the inorder traversal list. This function has a ...
My code: importjava.util.ArrayList;importjava.util.List;/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public List<Integer>preorderTraversal(TreeNode root...
Given therootof a binary tree, returnthe preorder traversal of its nodes' values. Example 1: image <pre>Input:root = [1,null,2,3] Output:[1,2,3] </pre> Example 2: <pre>Input:root = [] Output:[] </pre> Example 3:
102. 二叉树的层序遍历 - 给你二叉树的根节点 root ,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg] 输入:root = [3,9,20,null,null,15,7] 输出:[[3],[9,20],[15,7]]
每天一算:Binary Tree Preorder Traversal 程序员吴师兄 + 关注 预计阅读时间2分钟 6 年前 LeetCode上第144 号问题:二叉树的前序遍历 题目 给定一个二叉树,返回它的 前序 遍历。 示例: 输入: [1,null,2,3] 输出: [1,2,3] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题思路 用栈(Stack)...
TreeNode *t = stk.top(); rs.push_back(t->val); stk.pop(); flag = true; if (t->right) { stk.push(t->right); flag = false; } } return rs; } 或者如下,leetcode上解说的算法: vector<int> inorderTraversal(TreeNode *root) ...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<vector<int> > levelOrderBottom(TreeNode *root) { vector<vector<int>>res; vector<int>tmp; queue<TreeNode *>Qpre,Qcurr;if(root==NULL)returnres;else{ ...