Leetcode[105]-Construct Binary Tree from Preorder and Inorder Traversal

题目描述：给定前序和中序序列，构建一颗二叉树

节点定义为：

struct TreeNode{int val;TreeNode * left;TreeNode * right;TreeNode (int x) : val(x),left(NULL),right(NULL){}}; 

很明显，可以将大问题划分成子问题，用递归实现

时间复杂度为O(N)，空间复杂度为O(logN)

class Solution{public : TreeNode * construct(vector<int> & preorder,vector<int> & inorder){int len = preorder.size();if(len <= 0) return NULL;TreeNode * ans = createTree(preorder,0,len-1,inorder,0,len-1);return ans;}TreeNode * createTree(vector<int> & preorder,int prestart,int preend,vector<int> & inorder,int instart,int inend){if(instart > inend) return NULL;int index;for(int i = instart;i<=inend;i++){if(preorder[prestart] == inorder[i]){index = i;break;}}int lenth = index-instart;TreeNode * root = new TreeNode(preorder[prestart]);root->left = createTree(preorder,prestart+1,prestart+lenth,inorder,instart,index-1);root->right = createTree(preorder,prestart+lenth+1,preend,inorder,index+1,inend);return root;}};