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[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。
 1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public TreeNode buildTree(int[] preorder, int[] inorder) {12         return bulid(preorder,inorder, 0, 0,inorder.length-1);// p start of prorder ,ib start of inorder ,ie end of inorder ;13         14     }15 16 /**17  * Definition for binary tree18  * public class TreeNode {19  *     int val;20  *     TreeNode left;21  *     TreeNode right;22  *     TreeNode(int x) { val = x; }23  * }24  */25 26     public TreeNode bulid(int[]  preorder,int[] inorder,int p,int ib,int ie)27     {28       if(ib>ie) return null;29       int i; //split point30       for(i=ib;i<=ie;i++)31       {32           if(inorder[i]==preorder[p]) break;33       }34       TreeNode root=new TreeNode(preorder[p]);35      root.left= bulid(preorder,inorder,p+1,ib,i-1);36      root.right=bulid(preorder,inorder,p+i-ib+1,i+1,ie);//37       38        39     40            41       return root;42         43     }44 }

/**
* 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;

}

}