代码随想录算法训练营第十六天|104.二叉树的最大深度、559.n叉树的最大深度、111.二叉树的最小深度、222.完全二叉树的节点个数

今天是代码随想录算法训练营第十六天

今天写了4道力扣：分别是104.二叉树的最大深度、559.n叉树的最大深度、111.二叉树的最小深度、222.完全二叉树的节点个数

或许这里之后还要再看看吧。

104.二叉树的最大深度代码如下：

# 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 maxDepth(self, root: Optional[TreeNode]) -> int:
        return self.getdepth(root)

    def getdepth(self, node):
        if not node:
            return 0

        left_height = self.getdepth(node.left) # 得到左侧节点的高度
        right_height = self.getdepth(node.right) # 得到右侧节点的高度
        height = 1 + max(left_height, right_height) # 再考虑上中间节点的高度
        return height

559.n叉树的最大深度

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def maxDepth(self, root: 'Node') -> int:
        if not root: # 如果root为空，那么not root = True, 就会返回0
            return 0 
        
        max_depth = 1
        
        for child in root.children:
            max_depth = max(max_depth, self.maxDepth(child) + 1)
        
        return max_depth

111.二叉树的最小深度

# 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 getDepth(self, node):
        if node is None:
            return 0
        leftDepth = self.getDepth(node.left)  # 左
        rightDepth = self.getDepth(node.right)  # 右
        
        # 当一个左子树为空，右不为空，这时并不是最低点
        if node.left is None and node.right is not None:
            return 1 + rightDepth
        
        # 当一个右子树为空，左不为空，这时并不是最低点
        if node.left is not None and node.right is None:
            return 1 + leftDepth
        
        result = 1 + min(leftDepth, rightDepth)
        return result

    def minDepth(self, root):
        return self.getDepth(root)

222.完全二叉树的节点个数

# 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

# # 普通二叉树解法
# class Solution:
#     def countNodes(self, root: TreeNode) -> int:
#         return self.getNodesNum(root)
        
#     def getNodesNum(self, cur):
#         if not cur:
#             return 0
#         leftNum = self.getNodesNum(cur.left) #左
#         rightNum = self.getNodesNum(cur.right) #右
#         treeNum = leftNum + rightNum + 1 #中
#         return treeNum

# 完全二叉树解法
class Solution:
    def countNodes(self, root: TreeNode) -> int:
        if not root:
            return 0
        left = root.left
        right = root.right
        leftDepth = 0 #这里初始为0是有目的的，为了下面求指数方便
        rightDepth = 0
        while left: #求左子树深度
            left = left.left
            leftDepth += 1
        while right: #求右子树深度
            right = right.right
            rightDepth += 1
        if leftDepth == rightDepth:
            return (2 << leftDepth) - 1 #注意(2<<1) 相当于2^2，所以leftDepth初始为0
        return self.countNodes(root.left) + self.countNodes(root.right) + 1```