编写函数求二叉树的深度

算法思路：用递归实现该算法。如果二叉树为空，则返回0；如果二叉树只有一个结点（根结点），返回1，否则返回根结点的左分支的深度与右分支的深度中较大者加1。 based 《数据结构 C#语言版》

C# codes as below：

class Program

{

static void Main(string[] args)

{

TreeNode node1 = new TreeNode("1");

TreeNode node2 = new TreeNode("2");

TreeNode node3 = new TreeNode("3");

TreeNode node4 = new TreeNode("4");

TreeNode node5 = new TreeNode("5",node1,node2);

TreeNode node6 = new TreeNode("6", node5, node3);

TreeNode node7 = new TreeNode("7", node6);

TreeNode node8 = new TreeNode("8", node4, node7);

TreeNode node9 = new TreeNode("9", null, node8);

Console.WriteLine(DepthCount(node9));

Console.ReadKey();

}

static int DepthCount(TreeNode rootNode)

{

if (rootNode == null)

{

return 0;

}

else if (rootNode.LeftSon == null && rootNode.RightSon == null)

{

return 1;

}

else

{

int leftDepth = DepthCount(rootNode.LeftSon);

int rightDepth = DepthCount(rootNode.RightSon);

return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;

}

}

}

class TreeNode

{

public TreeNode LeftSon { get; set; }

public TreeNode RightSon { get; set; }

public string Value { get; set; }

public TreeNode(string value = null, TreeNode leftSon=null, TreeNode rightSon=null)

{

this.LeftSon = leftSon;

this.RightSon = rightSon;

this.Value = value;

}

}

output results:

6