[jzof] 07 rebuild binary tree

Here is a record of how the preorder cutoff subscript of the left subtree is calculated . It's convenient to recall later ：
Suppose this subscript is X. So there's the following equation ：
i - inStart = X - preStart;
therefore ：
X = preStart - inStart + i;
After this subscript is calculated , Other subscripts are simple .

import java.util.*;
/** * 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=reConstructBinaryTree(pre, 0, pre.length - 1, in, 0, in.length - 1);
        return root;
    }
    // The former sequence traversal {1,2,4,7,3,5,6,8} And middle order ergodic sequence {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);
                      break;
            }
                 
        return root;
    }
}