501. Find Mode in Binary Search Tree查找BST中的眾數
阿新 • • 發佈:2018-04-30
pan 為什麽 AR dup 不用 lan 結果 with bug
[抄題]:
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:
- The left subtree of a node contains only nodes with keys less than or equal to the node‘s key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node‘s key.
- Both the left and right subtrees must also be binary search trees.
For example:
Given BST [1,null,2,2]
,
1 2 / 2
return [2]
.
[暴力解法]:
時間分析:
空間分析:hashmap:n
[優化後]:
時間分析:
空間分析:各種count
[奇葩輸出條件]:
返回具體元素,不是次數。所以反過來 nums[次數] = 元素。
[奇葩corner case]:
[思維問題]:
新建數組指定空間,多大不知道,所以需要提前遍歷一下
[一句話思路]:
先存maxcount,符合要求之後再存modecount
[輸入量]:空: 正常情況:特大:特小:程序裏處理到的特殊情況:異常情況(不合法不合理的輸入):
[畫圖]:
[一刷]:
- 有返回值的單個函數中,必須把全局變量再寫一遍
- 要處理的情況寫if, else if,不處理的不用管
- inorder遍歷本質是dfs,也有退出條件
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分鐘肉眼debug的結果]:
[總結]:
[復雜度]:Time complexity: O() Space complexity: O()
[英文數據結構或算法,為什麽不用別的數據結構或算法]:
[關鍵模板化代碼]:
[其他解法]:
[Follow Up]:
[LC給出的題目變變變]:
[代碼風格] :
501. Find Mode in Binary Search Tree查找BST中的眾數