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。