第四题: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:

# -*- 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):
        if not pre or not tin:
            return None
        root = TreeNode(pre[0])
        root_id = tin.index(pre[0])
         
        root.left = self.reConstructBinaryTree(pre[1:root_id+1], tin[:root_id])
        root.right = self.reConstructBinaryTree(pre[root_id+1:], tin[root_id+1:])
         
        return root
        # write code here

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) {
        TreeNode root = helper(pre, 0, pre.length-1, in, 0, in.length-1);
        return root;
    }
    
    public TreeNode helper(int[] pre, int startPre, int endPre, int[] in, int startIn, int endIn){
        if(startPre > endPre || startIn > endIn){
            return null;
        }
        //获取当前子树的根节点，即前序遍历startPre指针指向的节点
        TreeNode root = new TreeNode(pre[startPre]);
        //在中序遍历中找到当前根节点所在位子，则其左边为其左子树的节点中序遍历结果，右边为其右子树的节点中序遍历结果
        for(int i=startIn; i<=endIn; i++){
            if(in[i] == root.val){
                //重构root的左子树
                root.left = helper(pre, startPre+1, startPre+i-startIn, in, startIn, i-1);
                //重构root的右子树
                root.right = helper(pre, startPre+i-startIn+1, endPre, in, i+1, endIn);
            }
        }
        
        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 reConstructBinaryTree(int [] pre,int [] in) {
        if((pre == null) || (in == null) || (pre.length==0) || (in.length==0)){
            return null;
        }
        TreeNode root = new TreeNode(pre[0]);
        int root_id = findIndex(in, pre[0]);
        
        root.left = reConstructBinaryTree(getSubArr(pre, 1, root_id+1), getSubArr(in, 0, root_id));
        root.right = reConstructBinaryTree(getSubArr(pre, root_id+1, pre.length), getSubArr(in, root_id+1, in.length));
        return root;
    }
    
    public int findIndex(int[] arr, int val){
        for(int i=0; i<arr.length; i++){
            if(arr[i] == val)
                return i;
        }
        return 0;
    }
    
    public int[] getSubArr(int[] arr, int start, int end){
        if((end > arr.length) || (start > end))
            return new int[0];
        int[] subArr = new int[end - start];
        for(int i=0; i < (end-start); i++){
            subArr[i] = arr[start+i];
        }
        return subArr;
    }
}