Construct Binary Tree from Preorder and Inorder Traversal

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

这道题是要根据先序遍历和中序遍历构造出一棵二叉树，这道题和上一题是一样的。上一题不过是通过后序遍历和中序遍历构造一棵二叉树。

只要将代码稍稍修改即可

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root;
        if(preorder.length == 0)
            return null;
        return createTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);

    }
    private TreeNode createTree(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart ,int inEnd){
        if(inStart > inEnd)
            return null;
        int value_root = preorder[preStart];                    //根据先序遍历第一个元素为根节点,找出根节点的值
        int index = 0;                                            //根节点在中序遍历中的索引值
        for(int i = inStart; i <= inEnd; i++){                    //查找根节点在中序遍历中的索引值
            if(value_root == inorder[i]){
                index = i;
                break;
            }
        }//for
        int length_leftTree = index - inStart;                    //左子树的长度
        TreeNode root_left = createTree(preorder, inorder, preStart + 1, preStart + length_leftTree, inStart, index - 1);
        TreeNode root_right = createTree(preorder, inorder, preStart + length_leftTree + 1, preEnd, index + 1, inEnd);
        
        TreeNode root = new TreeNode(value_root);
        root.left = root_left;
        root.right = root_right;
        
        return root;
    }
}