[LeetCode] 543. Diameter of Binary Tree

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.

这道题虽然是简单，但其实还是不简单。要求最长的路径，其实就是求根节点左右的最深路径相加，当然因为最长路径不一定经过根节点(如下图)，所以这个根节点的值要和左右的最大值分别比较。

          1
         / \
        2   3
       / \     
      4   5  
     /   / \
    6   7   8
             \
              9
               \
                10

有了这个基本的想法，那么解法一就是用分支的方法来求出根节点的周长，左边的最大周长，右边最大周长，然后取max，至于如何求最大周长，就要求该点的最大深度。但这个方法因为其实是用了两个recurive的方法来遍历了二叉树，所以时间复杂度是O(N²)

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

class Solution:
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        if not root: return 0
        pass_root = self.get_max_height(root.left) + self.get_max_height(root.right)
        return max(pass_root, self.diameterOfBinaryTree(root.left), self.diameterOfBinaryTree(root.right))
        
        
    def get_max_height(self, root: TreeNode) -> int:
        if not root: 
            return 0
        left = self.get_max_height(root.left)
        right = self.get_max_height(root.right)
        return max(left, right) + 1

有了上面的思路，我们可以进一步优化。发现其实可以用一个全局变量来记录每个根节点的最大的周长。这样用一个求最深路径的recursive就行了，时间复杂度优化到O(N)

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

class Solution:
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        self.res = 0
        self.get_max_height(root, self.res)
        return self.res
    
    def get_max_height(self, root: TreeNode, res: int) -> int:
        if not root: return 0
        
        left = self.get_max_height(root.left, self.res)
        right = self.get_max_height(root.right, self.res)
        self.res = max(self.res, left + right)
        #print(self.res)
        return max(left, right) + 1