leetcode 104. 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.

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

    3
   / 
  9  20
    /  
   15   7

return its depth = 3.

解法1:

# 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
        return max(self.maxDepth(root.left)+1, self.maxDepth(root.right)+1)
        

解法2: BFS

# 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
        q = [root]
        ans = 0
        while q:
            ans += 1
            q = [y for x in q for y in (x.left, x.right) if y]
        return ans           

DFS的迭代解法：在迭代tree node的过程中，使用另外一个stack追踪每一个node的高度。

# 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
        nodes = [root]
        heights = [1]        
        ans = 0
        while nodes:
            n = nodes.pop()
            h = heights.pop()
            ans = max(ans, h)
            if n.right:
                nodes.append(n.right)
                heights.append(h+1)
            if n.left:
                nodes.append(n.left)
                heights.append(h+1)            
        return ans