[leetcode] 652. Find Duplicate Subtrees @ python

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原题

Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any one of them.

Two trees are duplicate if they have the same structure with same node values.

Example 1:

    1
   / \
  2   3
 /   / \
4   2   4
   /
  4

The following are two duplicate subtrees:

  2
 /
4

and

4

Therefore, you need to return above trees’ root in the form of a list.

解法1

前序遍历+字典. 定义字典seen保存已见过的子树, 我们遍历二叉树, 用’#'代表空节点, 将每个节点对应的二叉树以字符串的形式存入字典, 如果seen中已经见过该二叉树, 则将节点加到res里, 这里注意当seen[string] == 1时才能往结果里加, 因为我们的结果不能有重复的节点.

代码

# 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 findDuplicateSubtrees(self, root):
        """
        :type root: TreeNode
        :rtype: List[TreeNode]
        """
        # pre-order
        def traverse(root):
            if not root:
                return '#'
            string = str(root.val) + traverse(root.left) + traverse(root.right)
            if string in seen and seen[string] == 1:
                res.append(root)
            seen[string] += 1
            return string
            
        res = []
        seen = collections.defaultdict(int)
        traverse(root)
        return res

解法2

后序遍历+字典. 思路与解法1相同, 我们从叶子节点往根节点读取二叉树.

代码

# 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 findDuplicateSubtrees(self, root):
        """
        :type root: TreeNode
        :rtype: List[TreeNode]
        """
        # post-order
        def traverse(root):
            if not root:
                return '#'
            string = traverse(root.left) + traverse(root.right) + str(root.val)
            if string in seen and seen[string] == 1:
                res.append(root)
            seen[string] += 1
            return string
            
        res = []
        seen = collections.defaultdict(int)
        traverse(root)
        return res