4、重建二叉树

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

=============Python=============

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        if len(inorder) == 0:
            return
        value = preorder.pop(0)
        ind = inorder.index(value)
        cur = TreeNode(value)
        cur.left = self.buildTree(preorder, inorder[:ind])
        cur.right = self.buildTree(preorder, inorder[ind+1:])
        return cur

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回构造的TreeNode根节点
    def reConstructBinaryTree(self, pre, tin):
        # write code here
        if not tin:
            return
        tmp = pre.pop(0)
        root = TreeNode(tmp)
        root.left = self.reConstructBinaryTree(pre, tin[0:tin.index(tmp)])
        root.right = self.reConstructBinaryTree(pre, tin[tin.index(tmp)+1:])
        return root

===============Java=================

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        if (pre == null || pre.length == 0 || in == null || in.length == 0 || pre.length != in.length) {
            return null;
        }
        return reConstruct(pre, in, 0, pre.length - 1, 0, in.length - 1);
    }
    
    public TreeNode reConstruct(int[] pre, int[] in, int preStart, int preEnd, int inStart, int inEnd) {
        if (preStart > preEnd || inStart > inEnd) {
            return null;
        }
        TreeNode head = new TreeNode(pre[preStart]);
        int inPos = 0;
        while (in[inPos] != pre[preStart]) {
            inPos++;
        }
        int minus = inPos - inStart;
        head.left = reConstruct(pre, in, preStart + 1, preStart + minus, inStart, inPos - 1);
        head.right = reConstruct(pre, in, preStart + minus + 1, preEnd, inPos + 1, inEnd);
        return head;
    }
}