一、题目：二叉树中和为某一值的路径
题目：输入一棵二叉树和一个整数，打印出二叉树中结点值的和为输入整数的所有路径。从树的根结点开始往下一直到叶结点所经过的结点形成一条路径。例如输入下图中二叉树和整数22，则打印出两条路径，第一条路径包含结点10、12，第二条路径包含结点10、5和7。
二叉树结点的定义如下：
public class BinaryTreeNode
{
public int Data { get; set; }
public BinaryTreeNode leftChild { get; set; }
public BinaryTreeNode rightChild { get; set; } public BinaryTreeNode(int data)
{
this.Data = data;
} public BinaryTreeNode(int data, BinaryTreeNode left, BinaryTreeNode right)
{
this.Data = data;
this.leftChild = left;
this.rightChild = right;
}
}
二、解题思路
2.1 核心步骤
首先，通过下图了解遍历上图中的二叉树的过程：
通过上图可以总结出规律：
（1）当用前序遍历的方式访问到某一结点时，我们把该结点添加到路径上，并累加该结点的值。
（2）如果该结点为叶结点并且路径中结点值的和刚好等于输入的整数，则当前的路径符合要求，我们把它打印出来。如果当前结点不是叶结点，则继续访问它的子结点。
（3）当前结点访问结束后，递归函数将自动回到它的父结点。这里要注意的是：在函数退出之前要在路径上删除当前结点并减去当前结点的值，以确保返回父结点时路径刚好是从根结点到父结点的路径。
2.2 代码实现
public static void FindPath(BinaryTreeNode root, int expectedSum)
{
if (root == null)
{
return;
} int currentSum = ;
List<int> path = new List<int>();
FindPath(root, expectedSum, path, ref currentSum);
} private static void FindPath(BinaryTreeNode root, int expectedSum, List<int> path, ref int currentSum)
{
currentSum += root.Data;
path.Add(root.Data);
// 如果是叶结点，并且路径上结点的和等于输入的值
// 打印出这条路径
bool isLeaf = root.leftChild == null && root.rightChild == null;
if (isLeaf && currentSum == expectedSum)
{
foreach (int data in path)
{
Console.Write("{0}\t", data);
}
Console.WriteLine();
} // 如果不是叶结点，则遍历它的子结点
if (root.leftChild != null)
{
FindPath(root.leftChild, expectedSum, path, ref currentSum);
} if (root.rightChild != null)
{
FindPath(root.rightChild, expectedSum, path, ref currentSum);
} // 在返回到父结点之前，在路径上删除当前结点，
// 并在currentSum中减去当前结点的值
path.Remove(root.Data);
currentSum -= root.Data;
}
三、单元测试
3.1 测试用例
（1）辅助方法的封装
private static void TestPortal(string testName, BinaryTreeNode root, int expectedSum)
{
if (!string.IsNullOrEmpty(testName))
{
Console.WriteLine("{0} begins:", testName);
} FindPath(root, expectedSum); Console.WriteLine();
} private static void SetSubTreeNode(BinaryTreeNode root, BinaryTreeNode lChild, BinaryTreeNode rChild)
{
if (root == null)
{
return;
} root.leftChild = lChild;
root.rightChild = rChild;
} private static void ClearUpTreeNode(BinaryTreeNode root)
{
if (root != null)
{
BinaryTreeNode left = root.leftChild;
BinaryTreeNode right = root.rightChild; root = null; ClearUpTreeNode(left);
ClearUpTreeNode(right);
}
}
（2）功能、特殊输入测试
// 10
// / \
// 5 12
// /\
// 4 7
// 有两条路径上的结点和为22
public static void Test1()
{
BinaryTreeNode node10 = new BinaryTreeNode();
BinaryTreeNode node5 = new BinaryTreeNode();
BinaryTreeNode node12 = new BinaryTreeNode();
BinaryTreeNode node4 = new BinaryTreeNode();
BinaryTreeNode node7 = new BinaryTreeNode(); SetSubTreeNode(node10, node5, node12);
SetSubTreeNode(node5, node4, node7); Console.WriteLine("Two paths should be found in Test1.");
TestPortal("Test1", node10, ); ClearUpTreeNode(node10);
} // 10
// / \
// 5 12
// /\
// 4 7
// 没有路径上的结点和为15
public static void Test2()
{
BinaryTreeNode node10 = new BinaryTreeNode();
BinaryTreeNode node5 = new BinaryTreeNode();
BinaryTreeNode node12 = new BinaryTreeNode();
BinaryTreeNode node4 = new BinaryTreeNode();
BinaryTreeNode node7 = new BinaryTreeNode(); SetSubTreeNode(node10, node5, node12);
SetSubTreeNode(node5, node4, node7); Console.WriteLine("No paths should be found in Test2.");
TestPortal("Test2", node10, ); ClearUpTreeNode(node10);
} // 5
// /
// 4
// /
// 3
// /
// 2
// /
// 1
// 有一条路径上面的结点和为15
public static void Test3()
{
BinaryTreeNode node5 = new BinaryTreeNode();
BinaryTreeNode node4 = new BinaryTreeNode();
BinaryTreeNode node3 = new BinaryTreeNode();
BinaryTreeNode node2 = new BinaryTreeNode();
BinaryTreeNode node1 = new BinaryTreeNode(); node5.leftChild = node4;
node4.leftChild = node3;
node3.leftChild = node2;
node2.leftChild = node1; Console.WriteLine("One path should be found in Test3.");
TestPortal("Test3", node5, ); ClearUpTreeNode(node5);
} // 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
// 没有路径上面的结点和为16
public static void Test4()
{
BinaryTreeNode node1 = new BinaryTreeNode();
BinaryTreeNode node2 = new BinaryTreeNode();
BinaryTreeNode node3 = new BinaryTreeNode();
BinaryTreeNode node4 = new BinaryTreeNode();
BinaryTreeNode node5 = new BinaryTreeNode(); node1.leftChild = node2;
node2.leftChild = node3;
node3.leftChild = node4;
node4.leftChild = node5; Console.WriteLine("No paths should be found in Test4.");
TestPortal("Test4", node1, ); ClearUpTreeNode(node1);
} // 树中只有1个结点
public static void Test5()
{
BinaryTreeNode node1 = new BinaryTreeNode(); Console.WriteLine("One paths should be found in Test5.");
TestPortal("Test5", node1, ); ClearUpTreeNode(node1);
} // 树中没有结点
public static void Test6()
{
Console.WriteLine("No paths should be found in Test6.");
TestPortal("Test6", null, );
}