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.通过给出的前序遍历与中序遍历还原二叉树。前序遍历：先访问根节点，然后左节点、右节点中序遍历：先访问左节点，然后根节点、右节点通过前序遍历找到每个子树的根节点，通过中序遍历找到每个子树的左右孩子范围*/// Definition for a binary tree node. #include <stdio.h> #include <iostream> #include <vector> #include <map> #include <hash_map> #include <unordered_map> using namespace std; //using namespace __gnu_cxx;  struct TreeNode {      int val;      TreeNode *left;      TreeNode *right;      TreeNode(int x) : val(x), left(NULL), right(NULL) {} };class Solution {public:    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {// 12ms        if(preorder.empty() || inorder.empty()) return NULL;        map<int,int> inorder_map;        for(int i=0;i<inorder.size();i++)//把中序遍历的值作为键，位置作为值存储在hash表中            inorder_map[inorder[i]]=i;        return constructTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1,inorder_map);    }    TreeNode* constructTree(vector<int>& preorder,int pl,int pr,vector<int>& inorder,int inl,int inr,map<int,int>& inorder_map)    {        if(pl>pr || inl>inr)            return NULL;        TreeNode *tree=new TreeNode(preorder[pl]);        int index=inorder_map[preorder[pl]];//根节点在中序遍历中的位置        tree->left=constructTree(preorder,pl+1,index-inl+pl,inorder,inl,index-1,inorder_map);        tree->right=constructTree(preorder,pl+index-inl+1,pr,inorder,index+1,inr,inorder_map);        return tree;    }    TreeNode* buildTree2(vector<int>& preorder, vector<int>& inorder)//29ms     {        if(preorder.empty() || inorder.empty()) return NULL;        return constructTree2(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);     }    TreeNode* constructTree2(vector<int>& preorder,int pl,int pr,vector<int>& inorder,int inl,int inr)    {        if(pl>pr || inl>inr) return NULL;        TreeNode* root=new TreeNode(preorder[pl]);        int index=inl;        for(;index<=inr;index++)//根节点在中序遍历中的位置            if(preorder[pl]==inorder[index]) break;        root->left=constructTree2(preorder,pl+1,index-inl+pl,inorder,inl,index-1);        root->right=constructTree2(preorder,pl+index-inl+1,pr,inorder,index+1,inr);        return root;    }};void printtree(struct TreeNode* root){    if(!root) return;    cout<<root->val<<endl;    printtree(root->left);    printtree(root->right);}int main(){    Solution mys;    vector<int> preorder={1,2,4,5,3,6,8,9};    vector<int> inorder={4,2,5,1,6,8,3,9};    struct TreeNode* root=mys.buildTree(preorder,inorder);    printtree(root);    return 0;}