构造树,并求每条路径和
第一步:构造树节点用到的类:
public class Node{ public int Val{get;set;} public Node? LNode{get;set;} public Node? RNode{get;set;} public Node(int val) { this.Val = val; } }
第二步构造树:
public static Node? BuildeTree(int[] arr,int li,int hi) { if (li > hi) return null; int mi = li + ((hi - li) >> 1); Node root = new Node(arr[mi]); root.LNode = BuildeTree(arr,li,mi-1); root.RNode = BuildeTree(arr,mi+1,hi); return root; }
第三步求路径和:
// 求路径和 递归 public static Dictionary<int,int> GetSumPath(Node root, int sumQ, Dictionary<int,int> nodeSum) { // 保护机制 if (root == null) return null; sumQ = sumQ + (root.Val); if (root.LNode == null && root.RNode == null) {nodeSum.Add(root.Val,sumQ);sumQ = sumQ - root.Val;} GetSumPath(root.LNode,sumQ,nodeSum); GetSumPath(root.RNode,sumQ,nodeSum); return nodeSum; }
运行代码:
int[] arr = {1,2,3,4,5,6,7}; var tree = TreeProject.BuildeTree(arr,0,arr.Length - 1); var dic = new Dictionary<int,int>(); TreeSum.GetSumPath(tree,0, dic); foreach(var item in dic) { System.Console.WriteLine( $"终结点名称:{item.Key} - 路径和:{item.Value } "); }
学习反思:
1.算法学习还是要画图,只靠脑子想是不行的
标签:Node,arr,01,之禅,递归,int,sumQ,root,public From: https://www.cnblogs.com/Insist-Y/p/17232470.html