标签：question oca statistic pac cti stack cat note red
Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.
Assume a BST is defined as follows:
For example:
Given BST [1,null,2,2],
1
2
/
2
return [2].
Note: If a tree has more than one mode, you can return them in any order.
Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).
粗暴解法，直接hash计数然后找出最大计数的值。
# 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 findMode(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
def dfs(node, cnt):
if not node: return
dfs(node.left, cnt)
cnt[node.val] += 1
dfs(node.right, cnt)
cnt = collections.defaultdict(int)
dfs(root, cnt)
ans,max_cnt = [],0
for k,v in cnt.items():
if v > max_cnt:
max_cnt = v
ans = [k]
elif v == max_cnt and k not in ans:
ans.append(k)
return ans
最后几行可以直接使用python max ：
mc = max(cnt.values()) return [n for n, c in cnt.items() if c == mc]
另外就是经典的tree node遍历解法，在dfs时候使用pre_node记录上次遍历的node，和当前node值进行比较：
# 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 findMode(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
ans,max_cnt = [],0
pre_node, pre_cnt = None, 1
def dfs(node):
nonlocal ans,max_cnt,pre_node,pre_cnt
if not node: return
dfs(node.left)
if not pre_node: # init
max_cnt = 1
ans = [node.val]
else:
if node.val == pre_node.val:
pre_cnt += 1
else:
pre_cnt = 1
if pre_cnt > max_cnt:
max_cnt = pre_cnt
ans = [node.val]
elif pre_cnt == max_cnt:
ans.append(node.val)
pre_node = node
dfs(node.right)
dfs(root)
return ans
python2 下的解法，合理运用dummy value其实非常方便哦！
class Solution(object):
def findMode(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root is None:
return []
self.curVal = root.val - 1 # dummy value is good!
self.curNum = 0 # dummy value is good!
self.maxNum = 0
self.maxVals = []
def dfs(root):
if root is not None:
dfs(root.right)
if root.val != self.curVal:
self.curNum = 0
self.curNum = self.curNum + 1
self.curVal = root.val
if self.curNum == self.maxNum:
self.maxVals.append(self.curVal)
elif self.curNum > self.maxNum:
self.maxNum = self.curNum
self.maxVals = [self.curVal]
dfs(root.left)
dfs(root)
return self.maxVals
使用stack的解法：
class Solution(object):
def findMode(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
stack, node, prev, cnt, res = [], root, None, 0, (0, [])
while stack or node:
if node:
stack.append(node)
node = node.left
else:
node = stack.pop()
if node.val != prev:
cnt = 0
cnt += 1
if cnt > res[0]:
res = (cnt, [node.val])
elif cnt == res[0]:
res[1].append(node.val)
prev = node.val
node = node.right
return res[1]
leetcode 501. Find Mode in Binary Search Tree
标签：question oca statistic pac cti stack cat note red
原文地址：https://www.cnblogs.com/bonelee/p/9147724.html
