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111. Minimum Depth of Binary Tree #22
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,50 @@ | ||
| - [111. Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree/description/) | ||
| - 方針 | ||
| - DFS | ||
| - 再帰関数 | ||
| - example. 2 で出ているような一直線タイプは前回の問題だと厳しいので、枝ノード(left があるのか?right があるのか?)の扱いに気をつけないといけない, 存在しないパスを最短と判断してしまうため | ||
| - 時間計算量: `O(N)`, 空間計算量: `O(N)` | ||
| - Stack | ||
| - 無理やりやれるけどやらない、前回の問題で[似たようなこと](https://github.com/hiroki-horiguchi-dev/leetcode/pull/21)をしているが、あまり勉強になるとは思わなかった | ||
| - BFS | ||
| - Queue | ||
| - 同様にやらない | ||
|
|
||
| ### 再帰関数 | ||
| ```java | ||
| /** | ||
| * Definition for a binary tree node. | ||
| * public class TreeNode { | ||
| * int val; | ||
| * TreeNode left; | ||
| * TreeNode right; | ||
| * TreeNode() {} | ||
| * TreeNode(int val) { this.val = val; } | ||
| * TreeNode(int val, TreeNode left, TreeNode right) { | ||
| * this.val = val; | ||
| * this.left = left; | ||
| * this.right = right; | ||
| * } | ||
| * } | ||
| */ | ||
| class Solution { | ||
| public int minDepth(TreeNode root) { | ||
| if (root == null) { | ||
| return 0; | ||
| } | ||
|
|
||
| int left = minDepth(root.left); | ||
| int right = minDepth(root.right); | ||
|
|
||
| if (root.left != null && root.right == null) { | ||
| return left + 1; | ||
| } | ||
|
|
||
| if (root.left == null && root.right != null) { | ||
| return right + 1; | ||
| } | ||
|
|
||
| return Math.min(left, right) + 1; | ||
| } | ||
| } | ||
| ``` | ||
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条件分岐は以下の方がわかりやすいと思いました。
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コメントの返信が遅くなりすみません。
ありがとうございます、おっしゃる通りですね。
方針部分で書いたように、leaf Node と brach node を分けて書くことを意識した結果、コメントいただいたコードになりました。
が、コメントいただいたコードの方が冗長な分岐と変数がなく、スッキリしていてわかりやすいですね。
「自然言語で方針を説明できる --> 手作業をそのまま起こしたようなコードを書く --> 冗長な箇所を削る」を含めて短時間でできるように意識していきたいと思います。