重建二叉树

题目

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。

思路及代码

核心思想：递归
前序遍历：根 左 右
中序遍历：左 根 右
左分支和右分支递归添加根节点

代码1

public class Solution {    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {        if(pre.length==0||in.length==0) //递归结束            return null;        TreeNode t = new TreeNode(pre[0]);        int left = 0, right = 0;        while (in[left] != pre[0]) {            left++;        }        right = in.length - left;        int[] in_left = new int[left];         int[] in_right = new int[right];        int[] pre_left = new int[left];         int[] pre_right = new int[right];        //左分支数组        for (int i = 0; i < left; i++) {            in_left[i] = in[i];            pre_left[i] = pre[i + 1];        }        t.left = reConstructBinaryTree(pre_left, in_left);        //右分支数组        for (int i = 0; i < right; i++) {            in_right[i] = in[i + left + 1];            pre_right[i] = pre[i + left + 1];        }        t.right = reConstructBinaryTree(pre_right, in_right);        return t;    }

代码2

链接：https://www.nowcoder.com/questionTerminal/8a19cbe657394eeaac2f6ea9b0f6fcf6来源：牛客网public class Solution {    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {        TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);        return root;    }    //前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}    private TreeNode reConstructBinaryTree(int [] pre,int startPre,int endPre,int [] in,int startIn,int endIn) {                 if(startPre>endPre||startIn>endIn)            return null;        TreeNode root=new TreeNode(pre[startPre]);                 for(int i=startIn;i<=endIn;i++)            if(in[i]==pre[startPre]){                root.left=reConstructBinaryTree(pre,startPre+1,startPre+i-startIn,in,startIn,i-1);                root.right=reConstructBinaryTree(pre,i-startIn+startPre+1,endPre,in,i+1,endIn);            }                         return root;    }}