利用二叉树的先序和中序（中序和后序）排列构建二叉树

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

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

/**
 * 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;
    }
};

/**
 * 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;
    }
};

  TreeNode Tnode(preorder[pStart]);

这里犯了一个很大的错误就是，其实每次算法在遇到该处时候只是将结构体中的成员变量替换成了新的，却没有构建新的节点。

为了验证我们的想法：

#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;
 }

}

最后是第二道类似的算法

/**
 * 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);
    }
};