原题地址

基本二叉树操作。

O[       ][              ]

[       ]O[              ]

代码:

 TreeNode *restore(vector<int> &preorder, vector<int> &inorder, int pp, int ip, int len) {
if (len <= )
return NULL; TreeNode *node = new TreeNode(preorder[pp]); if (len == )
return node; int leftLen = ;
while (inorder[ip + leftLen] != preorder[pp])
leftLen++;
node->left = restore(preorder, inorder, pp + , ip, leftLen);
node->right = restore(preorder, inorder, pp + leftLen + , ip + leftLen + , len - leftLen - ); return node;
} TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
return restore(preorder, inorder, , , preorder.size());
}

Leetcode#105 Construct Binary Tree from Preorder and Inorder Traversal的更多相关文章

  1. [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 ...

  2. (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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. [leetcode] 105. Construct Binary Tree from Preorder and Inorder Traversal (Medium)

    原题 题意: 根据先序和中序得到二叉树(假设无重复数字) 思路: 先手写一次转换过程,得到思路. 即从先序中遍历每个元素,(创建一个全局索引,指向当前遍历到的元素)在中序中找到该元素作为当前的root ...

  8. leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal,剑指offer 6 重建二叉树

    不用迭代器的代码 class Solution { public: TreeNode* reConstructBinaryTree(vector<int> pre,vector<in ...

  9. 【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 ...

随机推荐

  1. js 实现栈

    function Stack() { this.dataStore = []; this.top = 0; this.push=push; this.pop=pop; this.peek=peek; ...

  2. mariadb DML语句及用户授权

    DML(Data Manipulation Language):INSERT, DELETE, UPDATE, SELECT INSERT  [INTO]  tbl_name  [(col1,...) ...

  3. java枚举类型使用笔记

    1.values()方法返回枚举所有实例的一个数组,调用这个数组的length方法,可以得到这个枚举对象中实例的个数 2.枚举类的每个实例,其实都是static的,可以通过static方法直接调用,而 ...

  4. 计算系数 (codevs 1137) 题解

    [问题描述] 给定一个多项式(ax + by)^k,给定a.b.k.n.m,请求出多项式展开后x^n y^m项的系数. [样例输入] 1 1 3 1 2 [样例输出] 3 [解题思路] 本题为NOIP ...

  5. linux C 管道

    单一进程使用管道基本上毫无意义.管道一般用来子进程和父进程之间的通信,或者兄弟进程间的通信. 创建管道的主要函数是pipe #include<unistd.h> ]) pipe函数创建一个 ...

  6. C#串口操作类,包括串口读写操作

    串口进行操作的类,其中包括写和读操作,类可设置串口参数.设置接收函数.打开串口资源.关闭串口资源,操作完成后,一定要关闭串口.接收串口数据事件.接收数据出错事件.获取当前全部串口.把字节型转换成十六进 ...

  7. oracle - redo 损坏或删除处理方法

    OS: Oracle Linux Server release 5.7 DB: Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - ...

  8. 深入浅出const

    §通常,如果一个对象通过引用方式传到函数f中,而函数f又不会通过修改对象的数据成员的值改变该对象的状态,那么,我们最好将f的参数标记为const,这样可以预防对参数的误写,同时有些编译器还可对这种情况 ...

  9. Linux下编译内核配置选项简介

    Code maturity level options代码成熟度选项 Prompt for development and/or incomplete code/drivers 显示尚在开发中或尚未完 ...

  10. OpenGL 纹理贴图

    前一节实例代码中有个贴图操作. 今天就简单说明一下纹理贴图... 为了使用纹理贴图.我们首先需要启用纹理贴图功能. 我们可以在Renderer实现的onSurfaceCreated中定义启用: // ...