平衡二叉树路径和计算-优快云博客

本文链接：https://blog.youkuaiyun.com/lm393485/article/details/122129857

本文介绍了一种计算平衡二叉树所有路径和的方法，包括构建平衡二叉树、递归遍历树并计算从根节点到叶节点的所有路径之和。文中提供了一个具体的示例，展示了如何通过递归和非递归方式收集所有路径的和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

铁猛老师书学习笔记

<------------------------------------------------------------------------------------------>

问题1: int[]类型数组求和

问题2: stack的修改-递归-恢复

问题3: 平衡二叉树路径和

int[] SortArr = { 1, 2, 3, 4, 5, 6, 7};

Node Tree = BinaryTree(SortArr, 0, SortArr.Length - 1);

Console.WriteLine("Builde Tree Done!");


Queue<int> Sums = new Queue<int>();

GetPathSums(Tree, 0, Sums);

Console.WriteLine("Calculate Path-Sum Done!");


Console.Write("[");
foreach (var item in Sums.ToArray())
{
    Console.Write($"\t{item}");
}
Console.WriteLine("\t]");


List<int> list = CollectSums(Tree);
Console.WriteLine("Collect Path-Sum Done!");

foreach (var item in list)
{
    Console.Write($"{item} ");
}
Console.WriteLine();

/*----------------------------------------*/

static List<int> CollectSums(Node node)
{
    var sums = new List<int>();

    if (node == null)
    {
        return sums;
    }
    else
    {
        foreach (var item in CollectSums(node.Left))
        {
            sums.Add(item + (int)node.val);
        }

        foreach (var item in CollectSums(node.Right))
        {
            sums.Add(item + (int)node.val);
        }

        if (sums.Count == 0)
        {
            sums.Add((int)node.val);
        }

        return sums;
    }
    
}

static void GetPathSums(Node node, int ParentSum, Queue<int> sums)
{
    if (node == null)
    {
        return;
    }
    else
    {
        var CurSum = (int)node.val + ParentSum;

        if (node.Left == null && node.Right == null)
        {
            sums.Enqueue(CurSum);
        }

        GetPathSums(node.Left, CurSum, sums);
        GetPathSums(node.Right, CurSum, sums);
    }
}

static Node BinaryTree(int[] arr, int low, int max)
{
    if (low > max)
    {
        return null;
    }

    var mid = low + (max - low) / 2;

    var root = new Node(arr[mid]);

    root.Left = BinaryTree(arr, low, mid - 1);
    root.Right = BinaryTree(arr, mid + 1, max);


    return root;
}


public class Node
{
    public object val { get; set; }
    public Node Left { get; set; }
    public Node Right { get; set; }

    public Node(int val)
    {
        this.val = val;
        Left = null;
        Right = null;
    }
}

运行结果：