Leetcode_num2_Maximum Depth of Binary Tree

标签：leetcode   递归   

题目：

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.

AC率第二高的题啦，二项树的最长路径。初看此题就感觉要用递归，但不知怎的，一开始想到深度遍历上去了。。。。囧

实际上很简单的，某一节点的最长路径=max(该节点左子树的最长路径，该节点右子树的最长路径)+1

还有一点就是类中函数调用函数时格式为 self.函数名，否则会报 global name XXX is not defined

废话不多说啦，上代码咯

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return an integer
    def maxDepth(self, root):
        if root==None:
            return 0
        else:
            p=max(self.maxDepth(root.left),self.maxDepth(root.right))+1
            return p

Leetcode_num2_Maximum Depth of Binary Tree

标签：leetcode   递归   

原文地址：http://blog.csdn.net/eliza1130/article/details/39347315