算法系列15天速成 第十一天 树操作（上）

break;
                    case Direction.Right:
                        if (tree.right != null)
                            throw new Exception("树的右节点不为空，不能插入");
                        else
                            tree.right = node;

break;
                }
            }

BinTreeAddNode(tree.left, node, data, direction);
            BinTreeAddNode(tree.right, node, data, direction);

return tree;
        }
        #endregion

#region 获取二叉树指定孩子的状态
        /// <summary>
/// 获取二叉树指定孩子的状态
/// </summary>
/// <typeparam></typeparam>
/// <param></param>
/// <param></param>
/// <returns></returns>
        public ChainTree<T> BinTreeChild<T>(ChainTree<T> tree, Direction direction)
        {
            ChainTree<T> childNode = null;

if (tree == null)
                throw new Exception("二叉树为空");

switch (direction)
            {
                case Direction.Left:
                    childNode = tree.left;
                    break;
                case Direction.Right:
                    childNode = tree.right;
                    break;
            }

return childNode;
        }

#endregion

#region 获取二叉树的深度
        /// <summary>
/// 获取二叉树的深度
/// </summary>
/// <typeparam></typeparam>
/// <param></param>
/// <returns></returns>
        public int BinTreeLen<T>(ChainTree<T> tree)
        {
            int leftLength;
            int rightLength;

if (tree == null)
                return 0;

//递归左子树的深度
            leftLength = BinTreeLen(tree.left);

//递归右子书的深度
            rightLength = BinTreeLen(tree.right);

if (leftLength > rightLength)
                return leftLength + 1;
            else
                return rightLength + 1;
        }
        #endregion

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