BFS框架与终止条件-python-leetcode刷题

leetcode 树深搜索用BFS方法
104题最大深度 & 111题最小深度
区别只在于终止条件不同，BFS框架一致

搜最大深度时，即使碰到某个节点无，也继续搜

搜最小深度时，只要出现左右都无子节点的悬挂点，就返回深度

附代码
104最大深度

class Solution:
    def minDepth(self, root: TreeNode) -> int:
        # BFS解法--遍历每一层，有的话则depth+1
        # queue = collections.deque([root])
        queue = []
        queue.append(root)
        depth = 0

        while queue:
            size = len(queue)
            for _ in range(size):
                # cur = queue.popleft()
                cur = queue.pop(0)
                if not cur:
                    continue
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            depth += 1
        return depth

111最小深度

class Solution:
    def minDepth(self, root: TreeNode) -> int:
        # BFS解法二
        if not root:
            return 0
        
        depth = 1
        queue = []
        queue.append(root)
        while queue:
            for i in range(len(queue)):
                cur = queue.pop(0)
                if not cur.left and not cur.right:
                    return depth
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            depth += 1

题目：103. 二叉树的锯齿形层序遍历
难度：中等
算法：BFS

# 2022.09.21
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return[]
        queue = [root]
        ans = []
        index = 1   # 加个index
        while queue:
            level = []
            for i in range(len(queue)):
                node = queue.pop(0)
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            if level:
                if index == -1:      # 如果index为-1，就是偶数层
                    level = level[::-1]     # 偶数层翻一下
                ans.append(level)
                index = index * (-1)     # 奇数层偶数层来回标记
        return ans