104. Maximum Depth of Binary Tree

描述

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its depth = 3.

解法

DFS

深度优先的递归实现：

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if not root:
            return 0

        left = self.maxDepth(root.left)
        right = self.maxDepth(root.right)

        return 1 + max(left, right)

# or
class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right)) \
               if root else 0

BFS

广度优先需要借助队列实现，队列用于记录每一层的所有节点，每更新一层，深度就加一，当队列为空时说明不再有下一层节点，循环结束，返回深度。

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        depth = 0
        level = [root] if root else []

        while level:
            depth += 1
            queue = []  # named queue but not a real queue :)
            for el in level:
                if el.left:
                    queue.append(el.left)
                if el.right:
                    queue.append(el.right)
            level = queue

        return depth

# a real queue solution
 def maxDepth(self, root):
     if not root:
         return 0

     tqueue, h = collections.deque(), 0
     tqueue.append(root)
     while tqueue:
         nextlevel = collections.deque()
         while tqueue:
             front = tqueue.popleft()
             if front.left:
                 nextlevel.append(front.left)
             if front.right:
                 nextlevel.append(front.right)
         tqueue = nextlevel
         h += 1
     return h

# a stack solution
def maxDepth(self, root):
    if not root:
        return 0

    tstack,h = [root],0

    #count number of levels
    while tstack:
        nextlevel = []
        while tstack:
            top = tstack.pop()
            if top.left:
                nextlevel.append(top.left)
            if top.right:
                nextlevel.append(top.right)
        tstack = nextlevel
        h+=1
    return h