105. Construct Binary Tree from Preorder and Inorder Traversal

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Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return help(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
    }
    private TreeNode help(int[] preorder,int pre_start,int pre_end,int[] inorder,int in_start,int in_end){
        if(in_start>in_end || pre_start>pre_end) return null;
        
        
        TreeNode root = new TreeNode(preorder[pre_start]);
        int j = in_start;
        while(preorder[pre_start]!=inorder[j]&& j <=pre_end)
            j++;
        //注意下标！！！！！！！！
        root.left = help(preorder,pre_start+1,pre_start+(j-in_start),inorder,in_start,j-1);
        root.right = help(preorder,pre_start+(j-in_start)+1,pre_end,inorder,j+1,in_end);
        return root;
    }
}