Maximum Depth of Binary Tree

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.

Example

Given a binary tree as follow:

  1
 / \
2   3
   / \
  4   5

The maximum depth is3.

递归 树遍历的题最方便的写法自然是递归，不过递归调用的层数过多可能会导致栈空间溢出，因此需要适当考虑递归调用的层数。我们首先来看看使用递归如何解这道题，要求二叉树的最大深度，直观上来讲使用深度优先搜索判断左右子树的深度孰大孰小即可，从根节点往下一层树的深度即自增1，遇到NULL时即返回0。

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        return 1+max(self.maxDepth(root.left), self.maxDepth(root.right))

# 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 maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root is None:
            return 0
        left = self.maxDepth(root.left)
        right = self.maxDepth(root.right)
        return max(left, right)+1

由于对每个节点都会使用一次maxDepth，故时间复杂度为 O(n), 树的深度最大为 n, 最小为 log​2​​n, 故空间复杂度介于 O(logn) 和 O(n) 之间

非递归 ：level order

# 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 maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        level =[root]
        h = 0
        while level:
            nextlevel = []
            while level:
                node = level.pop()
                if node.left:
                    nextlevel.append(node.left)
                if node.right:
                    nextlevel.append(node.right)
            level = nextlevel
            h += 1
        return h