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.

思路：这道题的思路基本和上一题一样，不同之处在于后序遍历根节点在最后面。

/** * 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) {        return buildTree(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);    }    TreeNode* buildTree(vector<int>&inorder,int ileft,int iright,vector<int>&postorder,int pleft,int pright){        if(ileft>iright||pleft>pright) return NULL;        TreeNode *cur=new TreeNode (postorder[pright]);        int i=0;        for(i=ileft;i<=iright;i++){            if(inorder[i]==postorder[pright]) break;        }        cur->left=buildTree(inorder,ileft,i-1,postorder,pleft,pleft+i-ileft-1);        cur->right=buildTree(inorder,i+1,iright,postorder,pleft+i-ileft,pright-1);        return cur;    }};