[leedcode 106] 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.

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/*    注释见上一题：已知前序和中序，求树
关键点：
1 定位每层的根节点
2 计算好offset*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
if(inorder.length!=postorder.length||inorder.length==0) return null;
return getNode(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
}
TreeNode getNode(int[] inorder,int left1,int right1,int[] postorder,int left2,int right2){
if(left1>right1) return null;
if(left2>right2) return null;
int temp=postorder[right2];
int index=left1;
for(;index<=right1;index++){
if(inorder[index]==temp)break;
}
int len=index-left1;
TreeNode node=new TreeNode(temp);
node.left=getNode(inorder,left1,index-1,postorder,left2,left2+len-1);
node.right=getNode(inorder,index+1,right1,postorder,left2+len,right2-1);
return node;

}
}