利用二叉樹的先序和中序（中序和後序）排列構建二叉樹，二叉樹

利用二叉樹的先序和中序（中序和後序）排列構建二叉樹，二叉樹

題目來自於：

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

這一題目其實我想說的還不是My Code，是之前在寫代碼中遇到的一個bug問題。後面會進行詳細的解釋

Construct Binary Tree from Preorder and Inorder Traversal Total Accepted: 35628 Total Submissions: 134968My SubmissionsQuestion Solution 

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

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

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:TreeNode* buildhelp(vector<int>& preorder, vector<int>& inorder,int pStart,int pEnd,int iStart,int iEnd){      if(pStart>pEnd||iStart>iEnd)      return NULL;      TreeNode *Tnode=new TreeNode(preorder[pStart]);      if(pStart==pEnd)      return Tnode;      int cur=iStart;      while(inorder[cur]!=preorder[pStart])          cur++;  Tnode->left=buildhelp(preorder, inorder,pStart+1,pStart+(cur-iStart),iStart,cur-1);  Tnode->right=buildhelp(preorder, inorder,pEnd-(iEnd-cur-1),pEnd,cur+1,iEnd);      return Tnode;  }    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        TreeNode* Tree=NULL;        if(preorder.size()==0)         return Tree;        Tree=buildhelp(preorder, inorder,0,preorder.size()-1,0,inorder.size()-1);        return Tree;    }};

上述代碼是正確的可以ＡＣ的，但是本人在第一次寫的時候寫成了

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:TreeNode* buildhelp(vector<int>& preorder, vector<int>& inorder,int pStart,int pEnd,int iStart,int iEnd){      if(pStart<pEnd||iStart<iEnd)      return NULL;      TreeNode Tnode(preorder[pStart]);      if(pStart==pEnd)      return &Tnode;      int cur=iStart;      while(inorder[cur]!=preorder[pStart])          cur++;  Tnode.left=buildhelp(preorder, inorder,pStart+1,pStart+(cur-iStart),iStart,cur-1);  Tnode.right=buildhelp(preorder, inorder,pEnd-(iEnd-cur-1),pEnd,cur+1,iEnd);      return &Tnode;  }    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        TreeNode* Tree=NULL;        if(preorder.size()==0)         return Tree;        Tree=buildhelp(preorder, inorder,0,preorder.size()-1,0,inorder.size()-1);        return Tree;    }};

不同的地方在於

  TreeNode Tnode(preorder[pStart]);

然後返回其指標

這裡犯了一個很大的錯誤就是，其實每次演算法在遇到該處時候只是將結構體中的成員變數替換成了新的，卻沒有構建新的節點。

為了驗證我們的想法：

#include<iostream>#include<vector>using namespace std; struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;      TreeNode(int x) : val(x), left(NULL), right(NULL) {}  };int main(){int i=5; while(i--) {  TreeNode aa(i);  cout<<&aa<<endl; }}

我們在一個ｗｈｉｌｅ中多次重建該結構體但是其地址都沒有發生改變

最後是第二道類似的演算法

Construct Binary Tree from Inorder and Postorder Traversal Total Accepted: 32565 Total Submissions: 121341My SubmissionsQuestion Solution 

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

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

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:TreeNode* buildhelp(vector<int>& inorder, vector<int>& postorder,int iStart,int iEnd,int pStart,int pEnd){      if(pStart>pEnd||iStart>iEnd)      return NULL;      TreeNode Tnode(postorder[pEnd]);      if(pStart==pEnd)      return &Tnode;      int cur=iStart;      while(inorder[cur]!=postorder[pEnd])          cur++;  Tnode.left=buildhelp(inorder, postorder,iStart,cur-1,pStart,pStart+(cur-iStart-1));  Tnode.right=buildhelp(inorder, postorder,cur+1,iEnd,pEnd-1-(iEnd-cur-1),pEnd-1);      return &Tnode;  }    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {         if(inorder.size()==0)         return NULL;        return buildhelp(inorder, postorder,0,inorder.size()-1,0,postorder.size()-1);    }};