def preorer_elements(self):
"""先序遍历生成器"""
tree, stk = self._root, SStack()
while tree is not None or not stk.is_empty():
while tree is not None:
stk.push(tree.right)
yield tree.data
tree = tree.left
tree = stk.pop()
def DFS_graph(graph, v0):
"""从graph的v0顶点深有优先非递归得到的DFS序列"""
vnum = graph.vertex_num()
visited = [0] * vnum # 记录已访问顶点
visited[v0] = 1
DFS_seq = [v0] # 记录遍历序列
stk = SStack()
stk.push((0, graph.out_edges(v0))) # 入栈(i, edges)
yield v0
while not stk.is_empty(): # 下次应访问edges[i]
i, edges = stk.pop()
if i < len(edges):
v, _ = edges[i]
stk.push((i + 1, edges)) # 下次回来将访问edges[i+1]
if not visited[v]: # v未访问，访问并记录可达顶点
DFS_seq.append(v)
yield v # 返回生成器
visited[v] = 1
stk.push((0, graph.out_edges(v)))
def entries(self):
"""返回关键码、值对的生成器"""
tree, stk = self._root, SStack()
while tree is not None or not stk.is_empty():
while tree is not None:
stk.push(tree)
tree = tree.left
tree = stk.pop()
# 不直接返回tree.data（Assoc类对象），保证关键码的不可变
yield tree.data.key, tree.data.value
tree = tree.right
def values(self):
"""生成字典所有值序列的迭代器
中序遍历实现"""
tree, stk = self._root, SStack()
while tree is not None or not stk.is_empty():
while tree is not None:
stk.push(tree)
tree = tree.left
tree = stk.pop()
yield tree.data.value
tree = tree.right
def postorder_nonrec(tree, proc):
"""非递归后序遍历
1--内曾循环找当前子树的最下最左结点，将其入栈后终止
2--如果被访问结点是其父的左子结点，直接转到其右兄弟结点继续
3--如被处理结点是其父的右子结点，设tree为None将迫使外层循环的下次迭代弹出并访问更上一层的结点
"""
stk = SStack()
while tree or not stk.is_empty():
while tree: # 下行循环，直到栈顶的两子树空
stk.push(tree)
# 有左子树就取左入栈，因为有子树要先出栈
tree = tree.left if tree.left else tree.right
tree = stk.pop() # 访问栈顶
proc(tree.data)
if not stk.is_empty() and stk.top().left == tree: # 栈不空且当前结点是栈顶的左子节点
tree = stk.top().right
else:
tree = None # 没有右子树或右子树遍历完毕，强迫退栈
def preorder_elements(tree):
"""非递归先序遍历得到一个二叉树迭代器"""
stk = SStack()
while tree or not stk.is_empty():
while tree:
if tree.right is not None: # 空的右子树不入栈，节省栈空间
stk.push(tree.right) # 右子树入栈
yield tree.data
tree = tree.left # 处理左子树
if not stk.is_empty():
tree = stk.pop() # 左子树处理完，处理栈中的右子树
def preorder_nonrec(tree, proc):
"""非递归先序遍历
用栈缓存---先进后出---先将右子树压入栈"""
stk = SStack()
while tree or not stk.is_empty():
while tree:
if tree.right is not None: # 空的右子树不入栈，节省栈空间
stk.push(tree.right) # 右子树入栈
proc(tree.data)
tree = tree.left # 处理左子树
if not stk.is_empty():
tree = stk.pop() # 左子树处理完，处理栈中的右子树
def inorder_nonrec(tree, proc):
"""非递归中序遍历"""
stk = SStack()
while tree is not None or not stk.is_empty():
while tree:
stk.push(tree) # 不断压入左子树，root在栈底
tree = tree.left
tree = stk.pop()
proc(tree.data)
tree = tree.right # 处理右子树
def maze_solver(maze, start, end):
"""栈和回溯法"""
if start == end:
print(start)
return
st = SStack()
mark(maze, start)
st.push((start, 0)) # 入口和方向0
while not st.is_empty():
pos, nxt = st.pop() # 取栈顶及其探索方向
for i in range(nxt, 4): # 依次检查未探索方向
nextp = (pos[0] + dirs[i][0], pos[1] + dirs[i][1]) # 算出下一位置
if nextp == end:
# print_path(end, pos, st)
print(end, pos, end=' ')
while not st.is_empty():
print(st.pop()[0], end=" ")
return
if passable(maze, nextp): # 遇到未探索的新位置
st.push((pos, i + 1)) # 原位置和下一方向入栈
mark(maze, nextp)
st.push((nextp, 0)) # 新位置入栈
break # 退出内层循环， 此次迭代将以新栈顶为当前位置继续
print("No path found.")