[leetcode]重建二叉树（先序和终须） 中序遍和后续

分割后长度相等，就是参数麻烦，p，先序的起始点，  ib,ie 终须的结束和开始。

 /**
  * Definition for binary tree
  * 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 bulid(preorder,inorder, 0, 0,inorder.length-1);// p start of prorder ,ib start of inorder ,ie end of inorder ;

     }

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

     public TreeNode bulid(int[]  preorder,int[] inorder,int p,int ib,int ie)
     {
       if(ib>ie) return null;
       int i; //split point
       for(i=ib;i<=ie;i++)
       {
           if(inorder[i]==preorder[p]) break;
       }
       TreeNode root=new TreeNode(preorder[p]);
      root.left= bulid(preorder,inorder,p+1,ib,i-1);
      root.right=bulid(preorder,inorder,p+i-ib+1,i+1,ie);//

       return root;

     }
 }

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

}
public TreeNode bulid(int[] in,int[] pos,int p,int ib,int ie)
{
if(ib>ie) return null;

int i;
for(i=ib;i<=ie;i++)
{
if(pos[p]==in[i]) break;

}
TreeNode root=new TreeNode(pos[p]);
root.right=bulid(in,pos,p-1,i+1,ie);
root.left=bulid(in,pos,p-ie+i-1,ib,i-1);

return root;

}

}