feat: add solutions to lc problem: No.0314

No.0314.Binary Tree Vertical Order Traversal

Author: yanglbme

solution/0300-0399/0314.Binary Tree Vertical Order Traversal/README.md

@@ -46,7 +46,17 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-“BFS 层次遍历”实现。
+**方法一：DFS**
+
+DFS 遍历二叉树，记录每个节点的值、深度，以及横向的偏移量。然后对所有节点按照横向偏移量从小到大排序，再按照深度从小到大排序，最后按照横向偏移量分组。
+
+时间复杂度 $O(n\log \log n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。
+
+**方法二：BFS**
+
+本题较好的做法应该是 BFS，从上往下逐层进行遍历。
+
+时间复杂度 $O(n\log n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的结点数。
 
 <!-- tabs:start -->
 
@@ -62,22 +72,45 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def verticalOrder(self, root: TreeNode) -> List[List[int]]:
+    def verticalOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        def dfs(root, depth, offset):
+            if root is None:
+                return
+            d[offset].append((depth, root.val))
+            dfs(root.left, depth + 1, offset - 1)
+            dfs(root.right, depth + 1, offset + 1)
+
+        d = defaultdict(list)
+        dfs(root, 0, 0)
+        ans = []
+        for _, v in sorted(d.items()):
+            v.sort(key=lambda x: x[0])
+            ans.append([x[1] for x in v])
+        return ans
+```
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def verticalOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
         if root is None:
             return []
         q = deque([(root, 0)])
-        offset_vals = defaultdict(list)
+        d = defaultdict(list)
         while q:
-            node, offset = q.popleft()
-            offset_vals[offset].append(node.val)
-            if node.left:
-                q.append((node.left, offset - 1))
-            if node.right:
-                q.append((node.right, offset + 1))
-        res = []
-        for _, vals in sorted(offset_vals.items(), key=lambda x: x[0]):
-            res.append(vals)
-        return res
+            for _ in range(len(q)):
+                root, offset = q.popleft()
+                d[offset].append(root.val)
+                if root.left:
+                    q.append((root.left, offset - 1))
+                if root.right:
+                    q.append((root.right, offset + 1))
+        return [v for _, v in sorted(d.items())]
 ```
 
 ### **Java**
@@ -101,31 +134,247 @@ class Solution:
  * }
  */
 class Solution {
+    private TreeMap<Integer, List<int[]>> d = new TreeMap<>();
+
     public List<List<Integer>> verticalOrder(TreeNode root) {
+        dfs(root, 0, 0);
+        List<List<Integer>> ans = new ArrayList<>();
+        for (var v : d.values()) {
+            Collections.sort(v, (a, b) -> a[0] - b[0]);
+            List<Integer> t = new ArrayList<>();
+            for (var e : v) {
+                t.add(e[1]);
+            }
+            ans.add(t);
+        }
+        return ans;
+    }
+
+    private void dfs(TreeNode root, int depth, int offset) {
         if (root == null) {
-            return Collections.emptyList();
+            return;
         }
-        Map<Integer, List<Integer>> offsetVals = new TreeMap<>();
-        Map<TreeNode, Integer> nodeOffsets = new HashMap<>();
-        Deque<TreeNode> q = new ArrayDeque<>();
-        q.offer(root);
-        nodeOffsets.put(root, 0);
+        d.computeIfAbsent(offset, k -> new ArrayList<>()).add(new int[] {depth, root.val});
+        dfs(root.left, depth + 1, offset - 1);
+        dfs(root.right, depth + 1, offset + 1);
+    }
+}
+```
 
+```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 List<List<Integer>> verticalOrder(TreeNode root) {
+        List<List<Integer>> ans = new ArrayList<>();
+        if (root == null) {
+            return ans;
+        }
+        Deque<Pair<TreeNode, Integer>> q = new ArrayDeque<>();
+        q.offer(new Pair<>(root, 0));
+        TreeMap<Integer, List<Integer>> d = new TreeMap<>();
         while (!q.isEmpty()) {
-            TreeNode node = q.poll();
-            int offset = nodeOffsets.get(node);
-            offsetVals.computeIfAbsent(offset, k -> new ArrayList<>()).add(node.val);
-            if (node.left != null) {
-                q.offer(node.left);
-                nodeOffsets.put(node.left, offset - 1);
+            for (int n = q.size(); n > 0; --n) {
+                var p = q.pollFirst();
+                root = p.getKey();
+                int offset = p.getValue();
+                d.computeIfAbsent(offset, k -> new ArrayList()).add(root.val);
+                if (root.left != null) {
+                    q.offer(new Pair<>(root.left, offset - 1));
+                }
+                if (root.right != null) {
+                    q.offer(new Pair<>(root.right, offset + 1));
+                }
             }
-            if (node.right != null) {
-                q.offer(node.right);
-                nodeOffsets.put(node.right, offset + 1);
+        }
+        return new ArrayList<>(d.values());
+    }
+}
+```
+
+### **C++**
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+using pii = pair<int, int>;
+
+class Solution {
+public:
+    map<int, vector<pii>> d;
+
+    vector<vector<int>> verticalOrder(TreeNode* root) {
+        dfs(root, 0, 0);
+        vector<vector<int>> ans;
+        for (auto& [_, v] : d) {
+            sort(v.begin(), v.end(), [&](pii& a, pii& b) {
+                return a.first < b.first;
+            });
+            vector<int> t;
+            for (auto& x : v) {
+                t.push_back(x.second);
             }
+            ans.push_back(t);
         }
-        return new ArrayList<>(offsetVals.values());
+        return ans;
     }
+
+    void dfs(TreeNode* root, int depth, int offset) {
+        if (!root) return;
+        d[offset].push_back({depth, root->val});
+        dfs(root->left, depth + 1, offset - 1);
+        dfs(root->right, depth + 1, offset + 1);
+    }
+};
+```
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    vector<vector<int>> verticalOrder(TreeNode* root) {
+        vector<vector<int>> ans;
+        if (!root) return ans;
+        map<int, vector<int>> d;
+        queue<pair<TreeNode*, int>> q{{{root, 0}}};
+        while (!q.empty()) {
+            for (int n = q.size(); n; --n) {
+                auto p = q.front();
+                q.pop();
+                root = p.first;
+                int offset = p.second;
+                d[offset].push_back(root->val);
+                if (root->left) q.push({root->left, offset - 1});
+                if (root->right) q.push({root->right, offset + 1});
+            }
+        }
+        for (auto& [_, v] : d) {
+            ans.push_back(v);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func verticalOrder(root *TreeNode) [][]int {
+	d := map[int][][]int{}
+	var dfs func(*TreeNode, int, int)
+	dfs = func(root *TreeNode, depth, offset int) {
+		if root == nil {
+			return
+		}
+		d[offset] = append(d[offset], []int{depth, root.Val})
+		dfs(root.Left, depth+1, offset-1)
+		dfs(root.Right, depth+1, offset+1)
+	}
+	dfs(root, 0, 0)
+	idx := []int{}
+	for i := range d {
+		idx = append(idx, i)
+	}
+	sort.Ints(idx)
+	ans := [][]int{}
+	for _, i := range idx {
+		v := d[i]
+		sort.SliceStable(v, func(i, j int) bool { return v[i][0] < v[j][0] })
+		t := []int{}
+		for _, x := range v {
+			t = append(t, x[1])
+		}
+		ans = append(ans, t)
+	}
+	return ans
+}
+```
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func verticalOrder(root *TreeNode) [][]int {
+	ans := [][]int{}
+	if root == nil {
+		return ans
+	}
+	d := map[int][]int{}
+	q := []pair{pair{root, 0}}
+	for len(q) > 0 {
+		for n := len(q); n > 0; n-- {
+			p := q[0]
+			q = q[1:]
+			root = p.node
+			offset := p.offset
+			d[offset] = append(d[offset], root.Val)
+			if root.Left != nil {
+				q = append(q, pair{root.Left, offset - 1})
+			}
+			if root.Right != nil {
+				q = append(q, pair{root.Right, offset + 1})
+			}
+		}
+	}
+	idx := []int{}
+	for i := range d {
+		idx = append(idx, i)
+	}
+	sort.Ints(idx)
+	for _, i := range idx {
+		ans = append(ans, d[i])
+	}
+	return ans
+}
+
+type pair struct {
+	node   *TreeNode
+	offset int
 }
 ```