257. Binary Tree Paths返回所有深度優先的遍歷
[抄題]:
Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1 / 2 3 5
All root-to-leaf paths are:
["1->2->5", "1->3"]
[暴力解法]:
時間分析:
空間分析:
[奇葩corner case]:
[思維問題]:
以為這下需要匯總再返回了,其實不用,返回也是參數化了,在函數參數中進行的
[一句話思路]:
[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]:
[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] :
257. Binary Tree Paths返回所有深度優先的遍歷
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