【105 Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树】的更多相关文章

【LeetCode】105. Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树（Python）

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 日期 题目地址:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ 题目描述 Given preorder and inorder traversal of a tree, construct t…

105 Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树

给定一棵树的前序遍历与中序遍历,依据此构造二叉树.注意:你可以假设树中没有重复的元素.例如,给出前序遍历 = [3,9,20,15,7]中序遍历 = [9,3,15,20,7]返回如下的二叉树:    3   / \  9  20    /  \   15   7详见:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ Java实现: /** *…

LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal 由前序和中序遍历建立二叉树 C++

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [,,,,] inorder = [,,,,] Return the following binary tree: / \ / \ 前序.中序遍历得到二叉树,可以知道…

leetcode题解:Construct Binary Tree from Preorder and Inorder Traversal (根据前序和中序遍历构造二叉树)

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. 说明: 1)二叉树可空 2)思路:a.根据前序遍历的特点, 知前序序列(PreSequence)的首个元素(PreSequence[0])为二叉树的根(root),  然后在中序序列(InSequence)中查找此根(…

LeetCode：105_Construct Binary Tree from Preorder and Inorder Traversal | 根据前序和中序遍历构建二叉树 | Medium

要求:通过二叉树的前序和中序遍历序列构建一颗二叉树 代码如下: struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x): val(x),left(NULL), right(NULL) {} }; typedef vector<int>::iterator Iter; TreeNode *buildTree(vector<int> &preorder, vector<int&…

[Leetcode] Construct binary tree from preorder and inorder travesal 利用前序和中续遍历构造二叉树

Given preorder and inorder traversal of a tree, construct the binary tree. Note:  You may assume that duplicates do not exist in the tree. 利用前序和中序遍历构造二叉树的思想与利用中序和后续遍历的思想一样,不同的是,全树的根节点为preorder数组的第一个元素,具体分析过程参照利用中序和后续构造二叉树.这两道题是从二叉树的前序遍历.中序遍历.后续遍历这三种遍…

Leetcode：105. 从前序与中序遍历序列构造二叉树&106. 从中序与后序遍历序列构造二叉树

Leetcode:105. 从前序与中序遍历序列构造二叉树&106. 从中序与后序遍历序列构造二叉树 Leetcode:105. 从前序与中序遍历序列构造二叉树&106. 从中序与后序遍历序列构造二叉树 这道题是经典的模板题啦~ 用前序中序后序遍历结果建树的模板请跳转到:前中后序建立树或者直接历遍 直接默写!!! 105. 从前序与中序遍历序列构造二叉树 /** * Definition for a binary tree node. * struct TreeNode { * int v…

Java实现 LeetCode 105 从前序与中序遍历序列构造二叉树

105. 从前序与中序遍历序列构造二叉树 根据一棵树的前序遍历与中序遍历构造二叉树. 注意: 你可以假设树中没有重复的元素. 例如,给出 前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7] 返回如下的二叉树: 3 / \ 9 20 / \ 15 7 class Solution { private int pre=0; private int in=0; public TreeNode buildTree(int [] preor…

[LeetCode]105. 从前序与中序遍历序列构造二叉树(递归)、108. 将有序数组转换为二叉搜索树（递归、二分）

题目 05. 从前序与中序遍历序列构造二叉树 根据一棵树的前序遍历与中序遍历构造二叉树. 注意: 你可以假设树中没有重复的元素. 题解 使用HashMap记录当前子树根节点在中序遍历中的位置,方便每次查找. 递归 递归传参:该子树对应的前序遍历和中序遍历(用开始结束指针表示即可) 递归终止条件:序列长度为0,返回null: new当前子树根节点,左右孩子分别赋值为递归的返回值(通过确定左子树节点数确定前序遍历分割点),返回当前子树根节点. 代码 class Solution { private…

[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

【LeetCode】105. Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 可与Construct Binary Tree from Inorder and Postorder Trav…

【一天一道LeetCode】#105. Construct Binary Tree from Preorder and Inorder Traversal

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ Given preorder and inorder traversal of a tree, construct the bin…

LeetCode 中级 - 从前序与中序遍历序列构造二叉树(105)

一个前序遍历序列和一个中序遍历序列可以确定一颗唯一的二叉树. 根据前序遍历的特点, 知前序序列(PreSequence)的首个元素(PreSequence[0])为二叉树的根(root),  然后在中序序列(InSequence)中查找此根(root),  根据中序遍历特点, 知在查找到的根(root) 前边的序列为根的左子树的中序遍历序列,  后边的序列为根的右子树的中序遍历序列. 设在中序遍历序列(InSequence)根前边有left个元素. 则在前序序列(PreSequence)中, 紧…

LeetCode 105. 从前序与中序遍历序列构造二叉树（Construct Binary Tree from Preorder and Inorder Traversal）

题目描述 根据一棵树的前序遍历与中序遍历构造二叉树. 注意:你可以假设树中没有重复的元素. 例如,给出 前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7] 返回如下的二叉树: 3 / \ 9 20 / \ 15 7 解题思路 利用回溯的思想,分别记录生成树时前序遍历和中序遍历对应的段首.段尾,每次构造树时首先构造根节点为前序遍历的首节点,接着在中序遍历序列中找到根的位置,然后根左对应左子树,根右对应右子树,对应到前序遍历序列中分隔…

leetcode 105 106 从前序与中序遍历序列构造二叉树 从中序与后序遍历序列构造二叉树

题目: 105 根据一棵树的前序遍历与中序遍历构造二叉树. 注意:你可以假设树中没有重复的元素. 例如,给出 前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7] 返回如下的二叉树: 3 / \ 9 20 / \ 15 7 106 根据一棵树的中序遍历与后序遍历构造二叉树. 注意:你可以假设树中没有重复的元素. 例如,给出 中序遍历 inorder = [9,3,15,20,7] 后序遍历 postorder = [9,15,7,2…

LeetCode（105）：从前序与中序遍历序列构造二叉树

Medium! 题目描述: 根据一棵树的前序遍历与中序遍历构造二叉树. 注意:你可以假设树中没有重复的元素. 例如,给出 前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7] 返回如下的二叉树: 3 / \ 9 20 / \ 15 7 解题思路: 这道题要求用先序和中序遍历来建立二叉树,由于先序的顺序的第一个肯定是根,所以原二叉树的根节点可以知道,题目中给了一个很关键的条件就是树中没有相同元素,有了这个条件我们就可以在中序遍历中也定…

【leetcode 105. 从前序与中序遍历序列构造二叉树】解题报告

前往 中序,后序遍历构造二叉树, 中序,前序遍历构造二叉树 TreeNode* build(vector<int>& preorder, int l1, int r1, vector<int>& inorder, int l2, int r2) { if (l1>r1) return nullptr; ; // 确定当前根节点 for(i=l2;inorder[i]!=x&&i<r2;++i); // 在中序遍历序列中找到当前根节点位置(…

【LeetCode】105#从前序与中序遍历序列构造二叉树

题目描述 根据一棵树的前序遍历与中序遍历构造二叉树. 注意: 你可以假设树中没有重复的元素. 例如,给出 前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7] 返回如下的二叉树: 3 / \ 9 20 / \ 15 7 解题思路 首先列举一下二叉树的三种常见遍历方式: 前序遍历:根 - 左 - 右 中序遍历:左 - 根 - 右 后序遍历:左 - 右 - 根 可以发现,「前中后」是指根节点在三个节点中的相对顺序. 根据二叉树遍历的特点…

leetcode 105 Construct Binary Tree from Preorder and Inorder Traversal ----- java

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 给出前序遍历和中序遍历,然后求这棵树. 很有规律.递归就可以实现. /** * Definition for a binary tree node. * public class TreeNode { * int val; *…

105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. ============== 基本功: 利用前序和中序构建二叉树 , === code /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : va…

LeetCode OJ 105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 通过树的中序序列和前序序列来构建一棵树. 例如:一棵树的前序序列是1-2-3,中序序列有5种可能,分别是1-2-3.2-1-3.1-3-2.2-3-1.3-2-1.不同的中序序列对应着不同的树的结构,这几棵树分别是[1,nul…

【leetocde】 105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. Subscribe to see which companies asked this question     递归就可以了.   #include<algorithm> using namespace std; /**…

LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal （用先序和中序树遍历来建立二叉树）

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 题目标签:Array, Tree 题目给了我们preOrder 和 inOrder 两个遍历array,让我们建立二叉树.先来举一个例子,让我们看一下preOrder 和 inOrder的特性. / \   /      \…

(二叉树 递归) leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

105. Construct Binary Tree from Preorder and Inorder Traversal根据前中序数组恢复出原来的树

［抄题］: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \…

105. Construct Binary Tree from Preorder and Inorder Traversal (Tree; DFS)

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. class Solution { public: TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { root…

Java for LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 解题思路一: preorder[0]为root,以此分别划分出inorderLeft.preorderLeft.inorderRight.preorderRight四个数组,然后root.left=buildTree(pre…

[LC] 105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

LeetCodee 105. Construct Binary Tree from Preorder and Inorder Traversal

问题重述: 问题求解: 我们换一组比较有代表性的样例, 对于上图的树来说, index: 0 1 2 3 4 5 6 先序遍历为: 中序遍历为: 5同理,右子树也是如此.这样不难看出本题应该用递归方法解决. /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } *…

[Swift]LeetCode105. 从前序与中序遍历序列构造二叉树 | Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…