[easy+][Tree]543.

原题是：

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

Example:
Given a binary tree
1
/
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

思路是：

之前写过一个简单题，是求一个树的最大深度，这个深度不只是节点上的边，是指节点最深的路径上所有的节点数。这里可以用到这个函数。
最大diameter，必然是经过树或者他某个子树的root，所以这个路径是由某个节点的左右子树的最大深度的和构成。我们只要找到哪个子树拥有最大的过其root的路径就可以了。

代码是：

# 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 diameterOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None:
            return 0
        elif not (root.left or root.right):
            return 0
        
        ans = self.maxDepth(root.left) + self.maxDepth(root.right)
        Left = self.diameterOfBinaryTree(root.left)
        Right = self.diameterOfBinaryTree(root.right)
        res = max(max(ans,Left),Right)
        
        return res
        
    #return any tree's maximum depth(including root)
    def maxDepth(self,root): 
        if root == None:
            return 0
        
        Left = self.maxDepth(root.left)
        Right = self.maxDepth(root.right)
        return max(Left,Right) + 1