【leetcode刷题笔记】Construct Binary Tree from Preorder and Inorder Traversal

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

Note:
You may assume that duplicates do not exist in the tree.

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类似http://www.cnblogs.com/sunshineatnoon/p/3854935.html

只是子树的前序和中序遍历序列分别更新为：

//左子树：
left_prestart = prestart+1
left_preend = prestart+index-instart
//右子树
right_prestart = prestart+index-instart+1
right_preend =  preend

代码如下：

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int InorderIndex(int[] inorder,int key){
         if(inorder == null || inorder.length == 0)
             return -1;

         for(int i = 0;i < inorder.length;i++)
             if(inorder[i] == key)
                 return i;

         return -1;
     }
     public TreeNode buildTreeRec(int[] preoder,int[] inorder,int prestart,int preend,int instart,int inend){
         if(instart > inend)
             return null;
         TreeNode root = new TreeNode(preoder[prestart]);
         int index = InorderIndex(inorder, root.val);
         root.left = buildTreeRec(preoder, inorder, prestart+1, prestart+index-instart, instart, index-1);
         root.right = buildTreeRec(preoder, inorder, prestart+index-instart+1, preend, index+1, inend);

         return root;
     }
     public TreeNode buildTree(int[] preorder, int[] inorder) {
         return buildTreeRec(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);
     }
 }