【leetcode】106. Construct Binary Tree from Inorder and Postorder Traversal

题目要求：

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

给定二叉树后序和中序遍历，构造二叉树

思路：后序的最后一个结点是根节点，在中序遍历中找到对应的位置，记录左子树的size和右子树的size，递归构造左右子树

/**
* 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[] inorder, int[] postorder) {
if(inorder==null||postorder==null||postorder.length==0||inorder.length==0)
{
return null;

}
return buildWithPostIn(postorder,inorder,postorder.length-1,0,postorder.length);
}

//根据后序和中序构造二叉树
public TreeNode buildWithPostIn(int[] postorder,int[] inorder,int poststart,int instart,int size)
{

if(size<=0)
{
return null;
}
TreeNode root=new TreeNode(postorder[poststart]);
int index=0;
for(int i=0;i<size;i++)

{
if(postorder[poststart]==inorder[i+instart])
{
index=instart+i;
break;
}
}
int leftsize=index-instart;
int rightsize=size-leftsize-1;
root.left=buildWithPostIn(postorder,inorder,poststart-rightsize-1,instart,leftsize);
root.right=buildWithPostIn(postorder,inorder,poststart-1,index+1,rightsize);
return root;

}

}