牛客网 - 剑指Offer - 考点：树 题目：重建二叉树

题目描述

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。

/** * 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 rootNode = new TreeNode(pre[0]);                if (pre.length <= 1)        {            rootNode.left = null;            rootNode.right = null;                        return rootNode;        }                int NoRootIn = 0;                for (int i = 0; i < in.length; i++)        {            if (in[i] == rootNode.val)            {                NoRootIn = i;            }        }                if (NoRootIn > 0)        {            int [] preLeft = new int[NoRootIn];       int [] inLeft = new int[NoRootIn];                for (int i = 0; i < NoRootIn; i++)        {            preLeft[i] = pre[i+1];                inLeft[i] = in[i];        }            rootNode.left = reConstructBinaryTree(preLeft,inLeft);        }        else        {            rootNode.left = null;        }                if (pre.length - NoRootIn > 1)        {            int [] preRight = new int[pre.length - NoRootIn -1];       int [] inRight = new int[pre.length - NoRootIn -1];                for (int i = NoRootIn + 1; i < pre.length; i++)        {            preRight[i - NoRootIn - 1] = pre[i];                inRight[i - NoRootIn - 1] = in[i];        }            rootNode.right = reConstructBinaryTree(preRight,inRight);        }        else        {            rootNode.right = null;        }                return rootNode;    }}