Leetcode 144】的更多相关文章

144. Binary Tree Preorder Traversal Difficulty: Medium 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 it…
144. 二叉树的前序遍历 144. Binary Tree Preorder Traversal 题目描述 给定一个二叉树,返回它的 前序 遍历. LeetCode144. Binary Tree Preorder Traversal 示例: 输入: [1,null,2,3] 1 \ 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? Java 实现 Iterative Solution import java.util.LinkedList; import…
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? 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简…
144. 二叉树的前序遍历 给定一个二叉树,返回它的 前序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,2,3] /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { publi…
参考例子:[8,3,1,6,4,7,10,14,13] 8,3,1 和 6,4 说明从root开始,沿着左臂向下寻找leaf 的过程中应该逐个将node.val push入ans. class Solution(object): def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ stack = [] ans = [] while st…
方法一:(迭代) class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> v; if(root == NULL) return v; stack<TreeNode *> s; s.push(root); TreeNode *curNode = NULL; while(!s.empty()) { curNode = s.top(); s.pop(); v.pu…
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]. 二叉树的前序遍历,根节点→左子树→右子树 解题思路一: 递归实现,JAVA实现如下: public List<Integer> preorderTraversal(TreeNode root) { List<I…
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? 求前序遍历,要求不用递归.     使用双向队列.   /** * Definition…
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]. 解题思路: 这个问题最简单的方法是使用递归,但是题目规定不能使用,得使用迭代的方法. 那么我们考虑使用栈来实现. 思路是每次遍历这节点,把该点的值放入list中,然后把该点的右孩子放入栈中,并将当前点设置为左…
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? ---------------------------------------------------…