[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 tree 18 * 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 point 30 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;
}
}
