一、题目

给定一个有相同值的二叉搜索树（BST），找出 BST 中的所有众数（出现频率最高的元素）。

假定 BST 有如下定义：

结点左子树中所含结点的值小于等于当前结点的值
结点右子树中所含结点的值大于等于当前结点的值
左子树和右子树都是二叉搜索树

二、示例

例如：
给定 BST [1,null,2,2],

1
\
2
/
2
返回[2].

提示：如果众数超过1个，不需考虑输出顺序

进阶：你可以不使用额外的空间吗？（假设由递归产生的隐式调用栈的开销不被计算在内）

三、思路

1、中序遍历+有序数组的众数

2、遍历（先序，中序，后序均可）+dict字典

3、中序遍历

now记录上一个结点，now[0]为结点的val值，now[1]为该val值出现的次数；

max_val记录当前出现次数最大的值与次数，max_val[0]为次数最大值的val值，max_val[1]为次数最大值的次数；

四、代码

1、

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

class Solution:
    def findMode(self, root: TreeNode):
        # 中序遍历 + 有序数组求众数
        res = []
        def InTraverse(root):
            if root.left:
                InTraverse(root.left)
            res.append(root.val)
            if root.right:
                InTraverse(root.right)
        if root:
            InTraverse(root)
        # print(res)

        ans = []
        count = 0
        max_count = 0
        last = None

        # if len(res) == 0:
        #     return []

        for num in res:
            if last == num:
                count += 1
            else:
                count = 1
            if count > max_count:
                ans = [num]
                max_count = count
            elif count == max_count:
                ans.append(num)
            last = num
        return ans

if __name__ == '__main__':
    # root = TreeNode(1)
    # root.right = TreeNode(2)
    # root.right.left = TreeNode(2)
    # root.right.right = TreeNode(2)

    root = TreeNode(1)
    root.left = TreeNode(1)
    root.right = TreeNode(2)
    s = Solution()
    ans = s.findMode(root)
    print(ans)

2、

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

class Solution:
    def findMode(self, root: TreeNode):
        if root is None:
            return []
        dict = {}
        res = []

        def DFS(root):
            if root is None:
                return
            if root.val not in dict:
                dict[root.val] = 1
            else:
                dict[root.val] += 1
            DFS(root.left)
            DFS(root.right)
        DFS(root)
        # print(dict)
        max_val = max(dict.values())
        # print(max_val)
        for item, val in dict.items():
            if val == max_val:
                res.append(item)
        return res

if __name__ == '__main__':
    # root = TreeNode(1)
    # root.right = TreeNode(2)
    # root.right.left = TreeNode(2)
    # root.right.right = TreeNode(2)

    root = TreeNode(1)
    root.left = TreeNode(1)
    root.right = TreeNode(2)
    s = Solution()
    ans = s.findMode(root)
    print(ans)

3、

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

class Solution:
    def findMode(self, root: TreeNode):
        now = [0, 0]
        max_val = [0, 0]
        res = []

        def InTraverse(root):
            nonlocal now, max_val, res

            if root.left:
                InTraverse(root.left)

            if root.val == now[0]:
                now[1] += 1
            else:
                now[0] = root.val
                now[1] = 1

            if now[1] > max_val[1]:
                max_val[1] = now[1]
                max_val[0] = now[0]
                res = [max_val[0]]
            elif now[1] == max_val[1]:
                res.append(now[0])

            if root.right:
                InTraverse(root.right)

        if root:
            InTraverse(root)
        return res

if __name__ == '__main__':
    # root = TreeNode(1)
    # root.right = TreeNode(2)
    # root.right.left = TreeNode(2)
    # root.right.right = TreeNode(2)

    root = TreeNode(1)
    root.left = TreeNode(1)
    root.right = TreeNode(2)
    s = Solution()
    ans = s.findMode(root)
    print(ans)

LeetCode501：二叉搜索树中的众数

一、题目

二、示例

三、思路

四、代码

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