105. 从前序与中序遍历序列构造二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {//看笔记
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length>0&&inorder.length>0)
        {
            TreeNode root =new TreeNode(preorder[0]);//根节点

        
        int num=0;//根节点在中序序列里的下标
        for(int i=0;i<inorder.length;i++)
        {
            if(inorder[i]==root.val)
            {
                num=i;//num是下标，左子树加根节点总共num+1个，左子树num个元素
                break;
            }
        }
        int[] preLeft = Arrays.copyOfRange(preorder,1,num+1);//取数组里[a,b)区间的元素,这里是取了num个
        int[] preRight = Arrays.copyOfRange(preorder,num+1,preorder.length);//取出剩下的元素组成的数组，也就是右子树的元素的数组
        int[] inLeft=Arrays.copyOfRange(inorder,0,num);//取出中序遍历里面，左子树的元素组成的数组
        int[] inRight=Arrays.copyOfRange(inorder,num+1,inorder.length);
        root.left=buildTree(preLeft,inLeft);
        root.right=buildTree(preRight,inRight);
        return root;
        }
        else
        {
            return null;
        }
    }
}