• 剑指Offer面试题:23.二叉树中和为某一值的路径


    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 = 0;
            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}	", 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);
            }
        }
    View Code

      (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(10);
            BinaryTreeNode node5 = new BinaryTreeNode(5);
            BinaryTreeNode node12 = new BinaryTreeNode(12);
            BinaryTreeNode node4 = new BinaryTreeNode(4);
            BinaryTreeNode node7 = new BinaryTreeNode(7);
    
            SetSubTreeNode(node10, node5, node12);
            SetSubTreeNode(node5, node4, node7);
    
            Console.WriteLine("Two paths should be found in Test1.");
            TestPortal("Test1", node10, 22);
    
            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(10);
            BinaryTreeNode node5 = new BinaryTreeNode(5);
            BinaryTreeNode node12 = new BinaryTreeNode(12);
            BinaryTreeNode node4 = new BinaryTreeNode(4);
            BinaryTreeNode node7 = new BinaryTreeNode(7);
    
            SetSubTreeNode(node10, node5, node12);
            SetSubTreeNode(node5, node4, node7);
    
            Console.WriteLine("No paths should be found in Test2.");
            TestPortal("Test2", node10, 15);
    
            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(5);
            BinaryTreeNode node4 = new BinaryTreeNode(4);
            BinaryTreeNode node3 = new BinaryTreeNode(3);
            BinaryTreeNode node2 = new BinaryTreeNode(2);
            BinaryTreeNode node1 = new BinaryTreeNode(1);
    
            node5.leftChild = node4;
            node4.leftChild = node3;
            node3.leftChild = node2;
            node2.leftChild = node1;
    
            Console.WriteLine("One path should be found in Test3.");
            TestPortal("Test3", node5, 15);
    
            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(1);
            BinaryTreeNode node2 = new BinaryTreeNode(2);
            BinaryTreeNode node3 = new BinaryTreeNode(3);
            BinaryTreeNode node4 = new BinaryTreeNode(4);
            BinaryTreeNode node5 = new BinaryTreeNode(5);
    
            node1.leftChild = node2;
            node2.leftChild = node3;
            node3.leftChild = node4;
            node4.leftChild = node5;
    
            Console.WriteLine("No paths should be found in Test4.");
            TestPortal("Test4", node1, 16);
    
            ClearUpTreeNode(node1);
        }
    
        // 树中只有1个结点
        public static void Test5()
        {
            BinaryTreeNode node1 = new BinaryTreeNode(1);
    
            Console.WriteLine("One paths should be found in Test5.");
            TestPortal("Test5", node1, 1);
    
            ClearUpTreeNode(node1);
        }
    
        // 树中没有结点
        public static void Test6()
        {
            Console.WriteLine("No paths should be found in Test6.");
            TestPortal("Test6", null, 0);
        }

    3.2 测试结果

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  • 原文地址:https://www.cnblogs.com/edisonchou/p/4787736.html
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