[Force activity deduction 0310] 543. The binary tree diameter

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Title Description

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 Given a binary tree, you need to calculate the length of its diameter. A binary tree is the maximum length of the diameter of any two nodes in the path lengths. This path may pass through the root.

示例 :
给定二叉树

          1
         / \
        2   3
       / \     
      4   5    

返回 3, 它的长度是路径 [4,2,1,3] 或者 [5,2,1,3]。

注意：两结点之间的路径长度是以它们之间边的数目表示。

 

Problem-solving ideas 

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My thoughts

For two nodes longest path length, it requires only the number of edges in the left side of each node point + right, the maximum value for the iteration.

It simply is the most distant from the left to the far right path is the longest distance.

But such a problem would exist, each node requires a number of sides, there will be a large number of repetitive operations, what is the solution?

# 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
        """
        def layers(node):
            if not node:
                return 0
            return 1+max(layers(node.left),layers(node.right))
        if not root:
            return 0


        # return layers(root.right)
        stack = [root]
        ret = -1
        while len(stack):
            node = stack.pop()
            left , right = 0,0
            if node.left:
                left = layers(node.left)
                stack.append(node.left)
            if node.right:
                right = layers(node.right)
                stack.append(node.right)
            if left+right > ret:
                ret = left+right
        return ret

View Code

 

 

 

to sum up

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