Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

There is an example.
        _______7______
/ \
__10__ ___2
/ \ /
4 3 _8
\ /
1 11
The preorder and inorder traversals for the binary tree above is:
preorder = {7,10,4,3,1,2,8,11}
inorder = {4,10,3,1,7,11,8,2}

The first node in preorder alwasy the root of the tree. We can break the tree like:
1st round:
preorder:  {7}, {10,4,3,1}, {2,8,11}
inorder:     {4,10,3,1}, {7}, {11, 8,2}

        _______7______
/ \
{4,10,3,1} {11,8,2}
Since we alreay find that {7} will be the root, and in "inorder" sert, all the data in the left of {7} will construct the left sub-tree. And the right part will construct a right sub-tree. We can the left and right
part agin based on the preorder.

2nd round
left part                                                                            right part
preorder: {10}, {4}, {3,1}                                              {2}, {8,11}
inorder:  {4}, {10}, {3,1}                                                {11,8}, {2}

        _______7______
/ \
__10__ ___2
/ \ /
4 {3,1} {11,8}
see that, {10} will be the root of left-sub-tree and {2} will be the root of right-sub-tree.

Same way to split {3,1} and {11,8}, yo will get the complete tree now.

        _______7______
/ \
__10__ ___2
/ \ /
4 3 _8
\ /
1 11
So, simulate this process from bottom to top with recursion as following code.

c++

TreeNode *BuildTreePI(
vector<int> &preorder,
vector<int> &inorder,
int p_s, int p_e,
int i_s, int i_e){
if(p_s > p_e) return NULL;
int pivot = preorder[p_s];
int i = i_s;
for(;i<i_e;i++){
if(inorder[i] == pivot)
break;
}
int length1 = i-i_s-1;
int length2 = i_e-i-1;
TreeNode* node = new TreeNode(pivot);
node->left = BuildTreePI(preorder,inorder,p_s+1,length1+p_s+1,i_s, i-1);
node->right = BuildTreePI(preorder, inorder, p_e-length2, p_e, i+1, i_e);
return node;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
return BuildTreePI(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
}

java

public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildPI(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);
}
public TreeNode buildPI(int[] preorder, int[] inorder, int p_s, int p_e, int i_s, int i_e){
if(p_s>p_e)
return null;
int pivot = preorder[p_s];
int i = i_s;
for(;i<i_e;i++){
if(inorder[i]==pivot)
break;
}
TreeNode node = new TreeNode(pivot);
int lenLeft = i-i_s;
node.left = buildPI(preorder, inorder, p_s+1, p_s+lenLeft, i_s, i-1);
node.right = buildPI(preorder, inorder, p_s+lenLeft+1, p_e, i+1, i_e);
return node;
}

[LeetCode-21]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. 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 t ...

  7. 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 ...

  8. 【LeetCode】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 ...

  9. [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 ...

  10. Leetcode#105 Construct Binary Tree from Preorder and Inorder Traversal

    原题地址 基本二叉树操作. O[       ][              ] [       ]O[              ] 代码: TreeNode *restore(vector< ...

随机推荐

  1. Android布局中match_parent和fill_parent的差别

    今天在做项目的一个新功能的时候,从网上查找资源,发现android2.2中出现的MATCH_PARENT感到不明确.过去仅仅有FILL_PARENT和WRAP_CONTENT那么match_paren ...

  2. Mysql 没有nvl()函数,却有一个类似功能的函数ifnull()

    今天自己无聊写了看了一个查询需求随手写了一个sql语句,发现竟然不能运行,MySQL报[Err] 1305 - FUNCTION ceshi.nvl does not exist的错.才意识到自己写的 ...

  3. Atitit.导出excel报表的设计与实现java .net php 总

    Atitit.导出excel报表的设计与实现java .net php 总结 1. 导出报表 表格的设计要素1 1.1. 支持通用list<Map>转换1 1.2. 对于空列是否输出1 1 ...

  4. JavaScript之字符串、对象及操作符

    字符串-String 字符串就是字符序列. 字符串中,有些特殊字符,叫做字面量,常见的字面量如下表: 判断字符串长度使用length属性 text.length; 字符串拼接 var a = 'Jav ...

  5. JS DOM -- 关于回车键盘事件执行事件

    一.需求制作一个模拟对话框, 二. 1.需要发送后,输入框清空 2.按enter键可发送 三.代码部分 <!DOCTYPE HTML> <html> <head> ...

  6. redhat5.8下oracle11g启动失败

    # redhat5.8下oracle11g启动失败 ### 日志文件路径-----------------------------tail -f /u01/app/oracle/product/11. ...

  7. active mq 配置

    <transportConnectors> <!-- DOS protection, limit concurrent connections to 1000 and frame s ...

  8. 集合映射Set(使用xml文件)

    如果持久类具有Set对象,可以在映射文件中使用set元素映射Set集合. set元素不需要索引元素. List和Set之间的区别是: Set只存储唯一的值. 我们来看看我们如何在映射文件中实现集合: ...

  9. cocos2d-x:Layer::setPosition

    如果Node的实际类型是Layer或者其派生类, setPosition是不是有猫腻? std::string menuImage = "menu.png"; auto menuI ...

  10. lua(仿单继承)

    --lua仿单继承 Account = { balance = } function Account:new(o) o = o or {} setmetatable(o, self)--Account ...