105. 从前序与中序遍历序列构造二叉树

 public TreeNode buildTree(int[] preorder, int[] inorder) {
        return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }

    TreeNode build(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        if (preStart > preEnd) {
            return null;    //若先序序列长度小于等于0，则直接返回
        }
        int rootVal = preorder[preStart];   // root 节点对应的值就是前序遍历数组的第⼀个元素
        int index = 0;                      // rootVal 在中序遍历数组中的索引
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootVal) {
                index = i;
                break;
            }
        }
        int leftSize = index - inStart;
        TreeNode root = new TreeNode(rootVal);  // 先构造出当前根节点
        // 递归构造左右⼦树
        root.left = build(preorder, preStart + 1, preStart + leftSize, inorder, inStart, index - 1);
        root.right = build(preorder, preStart + leftSize + 1, preEnd, inorder, index + 1, inEnd);
        return root;
    }