[leedcode 105] 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.

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
//The preorder and inorder traversals for the binary tree above is:
/*   preorder = {7,10,4,3,1,2,8,11}
inorder = {4,10,3,1,7,11,8,2}

The first node in preorder alwasy the root of the tree. We can break the tree like:
1st round:
preorder:  {7}, {10,4,3,1}, {2,8,11}
inorder:     {4,10,3,1}, {7}, {11, 8,2}
可以发现，一趟遍历可以将数组一份为二，分别对应左子树集合和右子树集合
使用preorder数组定位跟节点，利用inorder数组分左右子树。
关键点：
1 定位每层的根节点
2 计算好offset
注意getTree函数的定义*/
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder.length!=inorder.length||preorder.length==0) return null;
return getTree(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
}
TreeNode getTree(int[] preorder,int left1,int right1,int[] inorder,int left2,int right2){
if(left1>right1) return null;
if(left2>right2) return null;
int temp=preorder[left1];
TreeNode node=new TreeNode(temp);
int index=left2;
for(;index<=right2;index++){
if(inorder[index]==temp)break;
}
int len=index-left2;
node.left=getTree(preorder,left1+1,left1+len,inorder,left2,index-1);
node.right=getTree(preorder,left1+len+1,right1,inorder,index+1,right2);
return node;
}
}