leetcode--Construct Binary Tree from Preorder and Inorder Traversal

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

Note:

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

分类：二叉树

题意：由前序遍历，中序遍历生成二叉树

解法1:递归。将前序遍历和中序遍历分成两部分。对于前序遍历而言，数组第一个元素为根节点。在中序遍历中找到这个根节点，其左右两边则分别是左右子树的中序遍历。

计算左子树长度len(根据根节点index求出)，然后对应于前序遍历，从根节点(数组第一个节点)开始,len长度之内，同样是左子树的前序遍历，剩下的是右子树的前序遍历。

这样我们又得到两个中序遍历和前序遍历，递归求解即可。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return hepler(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1);
}

public TreeNode hepler(int[] preorder,int pstart,int pend,int[] inorder,int istart,int iend){
if(pstart>pend) return null;
if(pstart==pend) return new TreeNode(preorder[pstart]);
int f = preorder[pstart];
TreeNode t = new TreeNode(f);
int mid = istart;
for(;mid<=iend;mid++){
if(f==inorder[mid]) break;
}
int len = mid-istart;
t.left = hepler(preorder, pstart+1, pstart+len, inorder, istart, mid-1);
t.right = hepler(preorder, pstart+len+1, pend, inorder, mid+1, iend);
return t;
}
}