求一个二叉树中任意两个节点间的距离，

网易有道笔试：
(1).
求一个二叉树中任意两个节点间的距离，
两个节点的距离的定义是 这两个节点间边的个数，
比如某个孩子节点和父节点间的距离是1，和相邻兄弟节点间的距离是2，优化时间空间复杂度。

C# codes as below:

using System;

namespace ConsoleApp

{

class RunClass

{

static void Main()

{

Node<int> i1 = new Node<int>() { Value = 4 };

Node<int> i2 = new Node<int>() { Value = 8 };

Node<int> i3 = new Node<int>() { Value = 12 };

Node<int> i4 = new Node<int>() { Value = 16 };

Node<int> i5 = new Node<int>() { Value = 6, Left = i1, Right = i2 };

Node<int> i6 = new Node<int>() { Value = 14, Left = i3, Right = i4 };

Node<int> i7 = new Node<int>() { Value = 10, Left = i5, Right = i6 };

Tree<int> tree = new Tree<int>() { Root = i7 };

int distance = new Helper().GetDistance<int>(tree, i4, i1);

Console.WriteLine(distance);

Console.ReadLine();

}

}

class Helper

{

public int GetDistance<T>(Tree<T> tree, Node<T> node1, Node<T> node2)

{

int length = (int)Math.Pow(2, CountMaxDepth<T>(tree) + 1) - 1;

Node<T>[] nodes = new Node<T>[length];

Node<T> root = tree.Root;

AddNodes<T>(nodes, root);

int index1 = FindNode<T>(nodes, node1);

int index2 = FindNode<T>(nodes, node2);

return CountDistance(index1, index2);

}

public int CountDistance(int i, int j)

{

if (i < j)

{

int k = i;

i = j;

j = k;

}

int distance = 0;

if (i != j)

{

distance += 1;

if (i % 2 == 1)

i = (i - 1) / 2;

else

i = i / 2 - 1;

distance += CountDistance(i, j);

}

return distance;

}

public int FindNode<T>(Node<T>[] nodes, Node<T> node)

{

for (int i = 0; i < nodes.Length; i++)

{

if (object.ReferenceEquals(nodes[i], node))

return i;

}

return -1;

}

private void AddNodes<T>(Node<T>[] nodes, Node<T> node, int currentIndex = 0)

{

nodes[currentIndex] = node;

if (node.Left != null)

{

AddNodes<T>(nodes, node.Left, 2 * currentIndex + 1);

}

if (node.Right != null)

{

AddNodes<T>(nodes, node.Right, 2 * currentIndex + 2);

}

}

public int CountMaxDepth<T>(Tree<T> tree)

{

if (tree == null||tree.Root==null)

return 0;

Node<T> root = tree.Root;

if (root.Left == null && root.Right == null)

return 0;

Tree<T> leftTree = new Tree<T>() { Root = root.Left };

Tree<T> rightTree = new Tree<T>() { Root = root.Right };

int leftMaxLevel = CountMaxDepth<T>(leftTree);

int rightMaxLevel = CountMaxDepth<T>(rightTree);

return leftMaxLevel > rightMaxLevel ? leftMaxLevel + 1 : rightMaxLevel + 1;

}

}

class Tree<T>

{

public Node<T> Root { get; set; }

}

class Node<T>

{

public Node<T> Left { get; set; }

public Node<T> Right { get; set; }

public T Value { get; set; }

}

}