经典：Construct Binary Tree from Preorder and Inorder Traversal：由先序中序构建二叉树

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

思路：手动模拟的时候，也是递归的重复。所以成序也是递归。为了减少重复时间，实现计算好中序中各个元素对应的下标位置。难点在于计算右子树时，需要计算偏移量。

这道题是必须会的一道题，非常非常的经典！

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {

public TreeNode dfs(int prestart,int instart,int inend,int[] preorder,int[] inorder,HashMap<Integer,Integer> map){
if(prestart>= preorder.length||instart >= inorder.length||instart > inend ) return null;
TreeNode n = new TreeNode(preorder[prestart]);
n.left = dfs(prestart+1,instart,map.get(preorder[prestart])-1,preorder,inorder,map);
n.right = dfs(prestart + map.get(preorder[prestart]) - instart +1,map.get(preorder[prestart]) + 1,inend,preorder ,inorder,map);
return n;
}

public TreeNode buildTree(int[] preorder, int[] inorder) {
HashMap map = new HashMap<Integer,Integer>();
for(int i = 0;i<inorder.length;i++){
map.put(inorder[i],i);
}
TreeNode root = dfs(0,0,inorder.length,preorder,inorder,map);
return root;

}
}