# 二叉树节点定义classTreeNode:def__init__(self,val):self.val=val self.left=None self.right=None # 二叉树的DFS遍历 defdfs_binary_tree(root):ifroot is None:returnprint(root.val,end=' ')dfs_binary_tree(root.left)dfs_binary_tree(root.right)# 构造二叉树 root=TreeNode(1)root.left=Tree...
lis.append(tree.val) return lis def BFS(self, tree): """ 广度优先遍历 层次遍历,lis是传入的list型参数,为了递归得到一个遍历List结果 """ ans = [] q1 = [tree] while q1: q2 = [] # q2 是q1 层的孩子,下一次循环的时候用它 for item in q1: if item: ans.append(item.val) q2.append...
void InOrder(BintreeNode<T> * subTree, void (* visit)(BintreeNode<T> *p)) { if (subTree != NULL) { InOrder(subTree->leftChild, visit); visit(subTree); InOrder(subTree->rightChild, visit); } }; // 递归方式实现后序遍历 void PostOrder(BintreeNode<T> *subTree, void (* visit)(...
图的一些相关概念: 简单图(Simple graph):无环并且无平行边的图. 路(path):内部点互不相同的链。 如果无向图G中每一对不同的顶点x和y都有一条路,(即W(G)=1,连通分支数)则称G是连通图,反之称为非连通图。 两端点相同的路(即闭路)称为圈(cycle)。 树(tree)是无圈连通无向图。树中度数为1的结点...
python代码: class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def pseudoPalindromicPaths (self, root: TreeNode): self.ass=[0,0,0,0,0,0,0,0,0,0] #计数路径中数字的个数 self.res=0 #记录...
DFS(path)#path="F:\\tree"path="F:\\AccessibilityServiceDemo"t=time.time() BFS(path) l1=time.time()-t t1=time.time() DFS(path) l2=time.time()-t1print(l1)print(l2) 另一种DFS写法: 将队列改为栈,先进后出 defDFS(start):
python class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def inorderTraversal(root): result = [] def dfs(node): if node: dfs(node.left) result.append(node.val) dfs(node.right) dfs(root) return result 在图...
python代码: class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def pseudoPalindromicPaths (self, root: TreeNode): self.ass=[0,0,0,0,0,0,0,0,0,0] #计数路径中数字的个数 self.res=0 #记录符...
以下是使用Python实现BFS实现树的层次遍历的示例代码：defbfs_tree_traversal(root):queue=[root]result=...
if succ not in disc: disc, result = visit(disc, succ) results.append(result) return results (对于上下文,这将是df-traverse函数的一部分,给定一个图和一个函数combiner :: Node_key -> [Result] -> Result,它将相当于构建depth-first林并在每个tree.上调用fold-tree combiner) ...