力扣热题100之二叉树的最大深度

题目

给定一个二叉树 root ，返回其最大深度。

二叉树的 最大深度 是指从根节点到最远叶子节点的最长路径上的节点数。

代码

方法一：递归

# 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:
        if not root:
            return 0
        leftdep=self.maxDepth(root.left)
        rightdep=self.maxDepth(root.right)
        return max(leftdep,rightdep)+1

方法二：栈迭代

# 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:
        if not root:
            return 0
        stack=[(root,1)]
        maxdepth=0
        while stack:
            node,depth=stack.pop()
            if node:
                maxdepth=max(depth,maxdepth)
                stack.append((node.left,depth+1))
                stack.append((node.right,depth+1))
        return maxdepth

方法三：队列迭代

# 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:
        if not root:
            return 0
        queue=deque([root])
        depth=0
        while queue:
            depth+=1# 每进入新的一层depth就加1
            for _ in range(len(queue)):#将下一层的节点加入队列
                node=queue.popleft()
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        return depth