python BFS和DFS LeetCode NO.102

python BFS和DFS LeetCode

BFS主要用队列来实现，DFS主要用栈来实现

#BFS模版
def BFS(graph,start,end):
	visited,quene = set(),[start]
	visited.add(start)
	while queue:
		node =quenue.pop()
		visited.add(node)
		process(node)
		nodes = generate_related_nodes(node)
		queuq.push(nodes)

#DFS不完全代码，用于面试模版，递归写法
visited = set()
def dfs (node, visited):
	visited.add(node)
	#对数据进行操作，
	...
	for next_node in node.children():
		if not next_node in visited:
					dfs(next_node,visited)

#非递归写法，类似于栈的结构
def DFS（self. tree):
	if tree.root is None:
		return []
	visited, stack = {},[tree.root]
	while stack:
		node= stack.pop()
		visited.add(node)
		#处理数据部分
		process(node)
		#查找子节点，或者是相互连接的节点（对于图而言）
		nodes = generate_related_nodes(node)
		stack.push(nodes)

如前文所述一样，BFS肯定是比较简单能想到的，用队列存储每一层的节点，然后一个一个的出队列，如果这个节点有孩子，那么就加入到队列中。当然也可以用DFS，DFS可以让面试官眼前一脸。

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root :
            return root
        if not root.left and not root.right:
            return [[root.val]]
        cur = [root]
        res = []
        while cur:
            nextStack,tmp = [],[]
            for node in cur:
                tmp.append(node.val)
                if node.left:
                    nextStack.append(node.left)
                if node.right:
                    nextStack.append(node.right)
            cur = nextStack
            res.append(tmp)
        return res

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root: return []
        
        result = []
        queue = collections.deque()
        queue.append(root)
        
        # visited = set(root)
        
        while queue:
            level_size = len(queue)
            current_level = []
            
            for _ in range(level_size):
                node = queue.popleft()
                current_level.append(node.val)
                if node.left: queue.append(node.left)
                if node.right:queue.append(node.right)
            
            result.append(current_level)
            
        return result

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root: return []
        self.result = []
        self._dfs(root,0)
        return self.result
    
    def _dfs(self, node, level):
        if not node:return 
        
        if len(self.result) <level + 1:
            self.result.append([])
            
        self.result[level].append(node.val)
        
        self._dfs(node.left, level+1)
        self._dfs(node.right,level+1)