1.题目描述 2.问题分析 利用递归. 3.代码 vector<int> preorderTraversal(TreeNode* root) { vector<int> v; preBorder(root, v); return v; } void preBorder(TreeNode *root, vector<int> &v) { if (root == NULL) return ; v.push_back(root->val); preBorder(…

Binary Tree Preorder Traversal 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? 解法一:递归 /** *…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 迭代 日期 题目地址:https://leetcode.com/problems/binary-tree-preorder-traversal/#/description 题目描述 Given a binary tree, return the preorder traversal of its nodes' values. For exampl…

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? 提示: 首先需要明确前序遍历的顺序,即:节点 -> 左孩子 -> 右孩子,这一…

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? Subscribe to see which companies asked this…

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]. 前序遍历二叉树,只不过题目的要求是尽量不要使用递归,当然我还是先用递归写了一个: /** * Definition for a binary tree node. * struct TreeNode { * int val…

Problem Link: http://oj.leetcode.com/problems/binary-tree-preorder-traversal/ Even iterative solution is easy, just use a stack storing the nodes not visited. Each iteration, pop a node and visited it, then push its right child and then left child in…

题目: Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 说明:1)下面有两种实现:递归(Recursive )与非递归(迭代iterati…

1.题目描述 2.问题分析 采用递归方法是标准解法. 3.代码 vector<int> preorder(Node* root) { vector<int> v; preNorder(root, v); return v; } void preNorder(Node *root , vector<int> &v) { if (root == NULL) return ; v.push_back(root->val); for (auto it = root…

1.题目描述 2.问题分析 递归 3.代码 vector<int> postorderTraversal(TreeNode* root) { vector<int> v; postBorder(root,v); return v; } void postBorder(TreeNode *root, vector<int> &v) { if (root == NULL) return ; postBorder(root->left, v); postBord…