Leetcode Construct Binary Tree from Inorder and Postorder Traversal

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

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

递归大法，代码如下：

/** * 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* buildTree(vector<int>& inorder, vector<int>& postorder) {        int plen = postorder.size(),ilen = inorder.size();                    return buildTree(postorder,0,plen-1,inorder,0,ilen-1);      }        TreeNode* buildTree(vector<int>& postorder,int pstart,int pend,vector<int>& inorder,int istart,int iend)      {          if(pstart > pend || istart > iend)              return NULL;                        TreeNode* root=new TreeNode(postorder[pend]);                    int i;          for(i=0;i<=iend-istart+1;i++)              if(postorder[pend] == inorder[istart+i])                  break;          root->left = buildTree(postorder,pstart,pstart+i-1,inorder,istart,istart+i-1);          root->right = buildTree(postorder,pstart+i,pend-1,inorder,istart+i+1,iend);          return root;      }  };

对于第二个buildtree函数中的那段for循环，这是最内层的，十分耗时，是否可以去掉？当然可以，借用hash结构map即可实现，代码如下：

/** * 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* buildTree(vector<int>& inorder, vector<int>& postorder) {        int plen = postorder.size(),ilen = inorder.size();          map<int,int> inMap;        for(int i=0;i<inorder.size();i++)            inMap[inorder[i]] = i;        return buildTree(postorder,0,plen-1,inorder,0,ilen-1,inMap);      }        TreeNode* buildTree(vector<int>& postorder,int pstart,int pend,vector<int>& inorder,int istart,int iend,map<int,int>& inMap)      {          if(pstart > pend || istart > iend)              return NULL;                        TreeNode* root=new TreeNode(postorder[pend]);                    int i;          i = inMap[postorder[pend]] - istart;                    root->left = buildTree(postorder,pstart,pstart+i-1,inorder,istart,istart+i-1,inMap);          root->right = buildTree(postorder,pstart+i,pend-1,inorder,istart+i+1,iend,inMap);          return root;      }  };

如果需要更快，那就需要迭代来代替递归。