Leetcode_num2_Maximum Depth of Binary Tree
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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 = Noneclass 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
补充更新ing~~~~
今天刷笔试题的时候又遇到了这道题,但是只能用c++来写,于是快速地写出了如下代码:
class Solution {public: int maxDepth(TreeNode *root) { if (root==NULL) return 0; else{ int rs=0; if(maxDepth(root->left)>maxDepth(root->right)){ rs=1+maxDepth(root->left); } else{ rs=1+maxDepth(root->right); } return rs; } }};
一运行,结果TLE了。。。。。囧
仔细检查发现该程序在判断和计算的过程中重复调用了递归函数,增加了算法复杂度,因此会出现TLE
修改后的代码如下:
class Solution {public: int maxDepth(TreeNode *root) { if (root==NULL) return 0; else{ int rs=0; int left=maxDepth(root->left); int right=maxDepth(root->right); if(left>right){ rs=1+left; } else{ rs=1+right; } return rs; } }};
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