[chjung99] WEEK 08 solutions

binary-tree-level-order-traversal/chjung99.java

@@ -0,0 +1,51 @@
+/**
+ * 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 List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> answer = new ArrayList<>();
+        Deque<Pair> queue = new ArrayDeque<>();
+        int[] visit = new int[2000];
+        if (root != null){
+            queue.add(new Pair(root, 0));
+        }
+
+        while (!queue.isEmpty()) {
+            Pair cur = queue.poll();
+            // System.out.println(cur.node.val+ " depth="+cur.depth);
+            if (visit[cur.depth] == 0){
+                answer.add(new ArrayList<>());
+                visit[cur.depth] = 1;
+            }
+            List<Integer> ls = answer.get(cur.depth);
+            ls.add(cur.node.val);
+
+            if (cur.node.left != null) queue.add(new Pair(cur.node.left, cur.depth + 1));
+            if (cur.node.right != null) queue.add(new Pair(cur.node.right, cur.depth + 1));
+        }
+        return answer;
+    }
+    class Pair{
+        TreeNode node;
+        int depth;
+
+        public Pair(TreeNode node, int depth){
+            this.node = node;
+            this.depth = depth;
+        }
+    }
+}
+
+

construct-binary-tree-from-preorder-and-inorder-traversal/chjung99.java

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+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+    // 빠른 조회를 위해 inorder의 값과 인덱스를 저장할 Map
+    private Map<Integer, Integer> inMap;
+
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        inMap = new HashMap<>();
+        for (int i = 0; i < inorder.length; i++) {
+            inMap.put(inorder[i], i);
+        }
+
+        return construct(preorder, 0, preorder.length - 1, 0, inorder.length - 1);
+    }
+
+    private TreeNode construct(int[] preorder, int preStart, int preEnd, int inStart, int inEnd) {
+        // 기저 조건: 더 이상 처리할 노드가 없는 경우
+        if (preStart > preEnd || inStart > inEnd) {
+            return null;
+        }
+
+        // 1. preorder의 현재 구간 첫 번째 원소가 루트 노드입니다.
+        int rootVal = preorder[preStart];
+        TreeNode root = new TreeNode(rootVal);
+
+        // 2. inorder에서 루트의 위치를 찾아 왼쪽/오른쪽 서브트리 범위를 나눕니다.
+        int rootIdx = inMap.get(rootVal);
+        int leftSize = rootIdx - inStart; // 왼쪽 서브트리에 포함된 노드 개수
+
+        // 3. 재귀적으로 왼쪽 자식 노드들을 연결합니다.
+        // preorder 범위: 루트 다음(preStart + 1)부터 개수(leftSize)만큼
+        // inorder 범위: 원래 시작점부터 루트 앞(rootIdx - 1)까지
+        root.left = construct(preorder, preStart + 1, preStart + leftSize,
+                inStart, rootIdx - 1);
+
+        // 4. 재귀적으로 오른쪽 자식 노드들을 연결합니다.
+        // preorder 범위: 왼쪽 식구들 끝난 지점 다음(preStart + leftSize + 1)부터 끝까지
+        // inorder 범위: 루트 다음(rootIdx + 1)부터 끝까지
+        root.right = construct(preorder, preStart + leftSize + 1, preEnd,
+                rootIdx + 1, inEnd);
+
+        return root;
+    }
+}
+
+

longest-consecutive-sequence/chjung99.java

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+class Solution {
+    int[] dp;
+    int[] next;
+    int n;
+    Map<Integer, Integer> location;
+
+    public int longestConsecutive(int[] nums) {
+        n = nums.length;
+        dp = new int[n];
+        next = new int[n];
+        location = new HashMap<>();
+
+        for (int i = 0; i < n; i++){
+            location.put(nums[i], i);
+            dp[i] = -1;
+        }
+
+        for (int i = 0; i < n; i++){
+            int nextValue = nums[i] + 1;
+            if (location.containsKey(nextValue)) {
+                next[i] = location.get(nextValue);
+                continue;
+            }
+            next[i] = -1;
+        }
+
+        for (int i = 0; i < n; i++) {
+            if (next[i] == -1) {
+                dp[i] = 1;
+                continue;
+            }
+            if (dp[next[i]] == -1) {
+                dp[i] = 1 + find(next[i], nums);
+                continue;
+            }
+            if (dp[next[i]] != -1){
+                dp[i] = 1 + dp[next[i]];
+            }
+        }
+        int answer = 0;
+        for (int i = 0; i < n; i++){
+            // System.out.print(dp[i] + " ");
+            answer = Math.max(answer, dp[i]);
+        }
+        return answer;
+    }
+
+    public int find(int idx, int[] nums) {
+        if (next[idx] == -1) {
+            dp[idx] = 1;
+        }
+        else if (dp[next[idx]] == -1) {
+            dp[idx] = 1 + find(next[idx], nums);
+        }
+        else if (dp[next[idx]] != -1) {
+            dp[idx] = 1 + dp[next[idx]];
+        }
+        return dp[idx];
+    }
+
+
+}
+
+

validate-binary-search-tree/chjung99.java

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+/**
+ * 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 {
+    boolean answer = true;
+    Deque<Integer> leftStack = new ArrayDeque<>();
+    Deque<Integer> rightStack = new ArrayDeque<>();
+
+    public boolean isValidBST(TreeNode root) {
+        rightStack.push(root.val);
+        dfs(root.left, root, "left");
+        rightStack.pop();
+
+        leftStack.push(root.val);
+        dfs(root.right, root, "right");
+        leftStack.pop();
+
+        return answer;
+    }
+
+    public void dfs(TreeNode node, TreeNode parent, String direction) {
+        if (node == null) return;
+
+        if (direction.equals("left") && node.val >= parent.val){
+            answer = false;
+        }
+        if (direction.equals("right") && node.val <= parent.val){
+            answer = false;
+        }
+        if (!leftStack.isEmpty() && node.val <= leftStack.peek()) {
+            answer = false;
+        }
+        if (!rightStack.isEmpty() && node.val >= rightStack.peek()) {
+            answer = false;
+        }
+
+        rightStack.push(node.val);
+        dfs(node.left, node, "left");
+        rightStack.pop();
+
+        leftStack.push(node.val);
+        dfs(node.right, node, "right");
+        leftStack.pop();
+    }
+}
+
+