week7 mission done

Author: dev-jonghoonpark

binary-tree-level-order-traversal/dev-jonghoonpark.md

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+- 문제: https://leetcode.com/problems/binary-tree-level-order-traversal/
+- time complexity : O(n)
+- space complexity : O(n), 트리가 균등한 형태일 경우 O(logn)에 가까워진다. (결과에 사용되는 list는 고려하지 않는다.)
+- 블로그 링크 : https://algorithm.jonghoonpark.com/2024/06/10/leetcode-102
+
+```java
+class Solution {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+
+        dfs(result, root, 0);
+
+        return result;
+    }
+
+    public void dfs(List<List<Integer>> result, TreeNode node, int level) {
+        if(node == null) {
+            return;
+        }
+
+        if (result.size() <= level) {
+            result.add(new ArrayList<>());
+        }
+        result.get(level).add(node.val);
+
+        dfs(result, node.left, level + 1);
+        dfs(result, node.right, level + 1);
+    }
+}
+```

lowest-common-ancestor-of-a-binary-search-tree/dev-jonghoonpark.md

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+- 문제: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
+- time complexity : O(n)
+- space complexity : O(n), 트리가 균등한 형태일 경우 O(logn)에 가까워진다.
+- 블로그 링크 : https://algorithm.jonghoonpark.com/2024/06/10/leetcode-235
+
+```java
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        return dfs(root, p, q);
+    }
+
+    private TreeNode dfs(TreeNode node, TreeNode p, TreeNode q) {
+        if (node == null) {
+            return null;
+        }
+
+        TreeNode left = dfs(node.left, p, q);
+        TreeNode right = dfs(node.right, p, q);
+
+        if ((left != null && right != null) || node.val == p.val || node.val == q.val) {
+            return node;
+        }
+
+        return left != null ? left : right;
+    }
+}
+```

remove-nth-node-from-end-of-list/dev-jonghoonpark.md

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+- 문제: https://leetcode.com/problems/remove-nth-node-from-end-of-list/
+- time complexity : O(n)
+- space complexity : O(1)
+- 블로그 링크 : https://algorithm.jonghoonpark.com/2024/06/10/leetcode-19
+
+```java
+class Solution {
+    public ListNode removeNthFromEnd(ListNode head, int n) {
+        ListNode current = head;
+        int length = 0;
+        while(current != null) {
+            length++;
+            current = current.next;
+        }
+
+        if (length == 1) {
+            return null;
+        }
+
+        if (length == n) {
+            return head.next;
+        }
+
+        int removeIndex = length - n;
+        int pointer = 0;
+        current = head;
+        while (pointer < removeIndex - 1) {
+            current = current.next;
+            pointer++;
+        }
+        current.next = current.next.next;
+        return head;
+    }
+}
+```

reorder-list/dev-jonghoonpark.md

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+- 문제 : https://leetcode.com/problems/reorder-list/
+- 블로그 링크 : https://algorithm.jonghoonpark.com/2024/06/10/leetcode-143
+
+## 방법 1 (list 사용)
+
+- time complexity : O(n)
+- space complexity : O(n)
+
+```java
+class Solution {
+    public void reorderList(ListNode head) {
+        List<ListNode> list = new ArrayList<>();
+
+        ListNode current = head;
+        while (current.next != null) {
+            list.add(current);
+            current = current.next;
+        }
+        list.add(current);
+
+        ListNode reordered = head;
+        for (int i = 1; i < list.size(); i++) {
+            if (i % 2 == 0) {
+                reordered.next = list.get(i / 2);
+            } else {
+                reordered.next = list.get(list.size() - ((i / 2) + 1));
+            }
+            reordered = reordered.next;
+        }
+        reordered.next = null;
+    }
+}
+```
+
+## 방법 2 (pointer 사용)
+
+- time complexity : O(n)
+- space complexity : O(1)
+
+```java
+class Solution {
+    public void reorderList(ListNode head) {
+        ListNode prev = null;
+        ListNode slow = head;
+        ListNode fast = head;
+
+        while(slow != null && fast != null && fast.next != null) {
+            prev = slow;
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+
+        if (prev == null) {
+            return;
+        }
+
+        prev.next = null;
+
+        ListNode current = head;
+        ListNode reversed = reverse(slow);
+        while(true) {
+            ListNode temp = current.next;
+
+            if (reversed != null) {
+                current.next = reversed;
+                reversed = reversed.next;
+                current = current.next;
+            }
+
+            if (temp != null) {
+                current.next = temp;
+                current = current.next;
+            } else {
+                current.next = reversed;
+                break;
+            }
+        }
+    }
+
+    public ListNode reverse(ListNode treeNode) {
+        ListNode current = treeNode;
+        ListNode prev = null;
+        while(current != null) {
+            ListNode temp = current.next;
+            current.next = prev;
+            prev = current;
+            current = temp;
+        }
+        return prev;
+    }
+}
+```

validate-binary-search-tree/dev-jonghoonpark.md

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+- 문제 : https://leetcode.com/problems/validate-binary-search-tree/
+- time complexity : O(n)
+- space complexity : O(n), 트리가 균등한 형태일 경우 O(logn)에 가까워진다.
+- 블로그 링크 : https://algorithm.jonghoonpark.com/2024/06/10/leetcode-98
+
+```java
+class Solution {
+    public boolean isValidBST(TreeNode root) {
+        return dfs(root.left, (long) Integer.MIN_VALUE - 1, root.val)
+               && dfs(root.right, root.val, (long) Integer.MAX_VALUE + 1);
+    }
+
+    private boolean dfs(TreeNode node, long min, long max) {
+        if (node == null) {
+            return true;
+        }
+
+        // left로 갈 때 max를 자신으로, right로 갈 때는 min을 자신으로
+        if (!(dfs(node.left, min, node.val)
+              && dfs(node.right, node.val, max))) {
+            return false;
+        }
+
+        return node.val > min && node.val < max;
+    }
+}
+```