feat: add solutions to lc problem: No.0662

No.0662.Maximum Width of Binary Tree

Author: yanglbme

solution/0600-0699/0662.Maximum Width of Binary Tree/README.md

@@ -57,7 +57,21 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-BFS 层序遍历。
+**方法一：BFS**
+
+对节点进行编号，初始根节点编号为 $1$。
+
+对于一个编号为 `i` 的节点，它的左节点编号为 `i<<1`，右节点编号为 `i<<1|1`。
+
+采用 BFS 进行层序遍历，求每层的宽度时，用该层的最大节点编号减去最小节点编号再加一即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。
+
+**方法二：DFS**
+
+定义 `dfs(root, depth, i)` 表示从深度为 `depth`，且编号为 `i` 的节点 `root` 开始往下搜索。记录每一层最先访问到的节点的编号。访问到当前层其它节点时，求当前节点编号与当前层最小编号的差再加一，更新当前层的最大宽度。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。
 
 <!-- tabs:start -->
 
@@ -73,18 +87,43 @@ BFS 层序遍历。
 #         self.left = left
 #         self.right = right
 class Solution:
-    def widthOfBinaryTree(self, root: TreeNode) -> int:
-        q = deque([(root, 1)])
+    def widthOfBinaryTree(self, root: Optional[TreeNode]) -> int:
         ans = 0
+        q = deque([(root, 1)])
         while q:
-            n = len(q)
             ans = max(ans, q[-1][1] - q[0][1] + 1)
-            for _ in range(n):
-                node, j = q.popleft()
-                if node.left:
-                    q.append((node.left, 2 * j))
-                if node.right:
-                    q.append((node.right, 2 * j + 1))
+            for _ in range(len(q)):
+                root, i = q.popleft()
+                if root.left:
+                    q.append((root.left, i << 1))
+                if root.right:
+                    q.append((root.right, i << 1 | 1))
+        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 widthOfBinaryTree(self, root: Optional[TreeNode]) -> int:
+        def dfs(root, depth, i):
+            if root is None:
+                return
+            if len(t) == depth:
+                t.append(i)
+            else:
+                nonlocal ans
+                ans = max(ans, i - t[depth] + 1)
+            dfs(root.left, depth + 1, i << 1)
+            dfs(root.right, depth + 1, i << 1 | 1)
+
+        ans = 1
+        t = []
+        dfs(root, 0, 1)
         return ans
 ```
 
@@ -111,17 +150,19 @@ class Solution:
 class Solution {
     public int widthOfBinaryTree(TreeNode root) {
         Deque<Pair<TreeNode, Integer>> q = new ArrayDeque<>();
-        q.offerLast(new Pair<>(root, 1));
+        q.offer(new Pair<>(root, 1));
         int ans = 0;
         while (!q.isEmpty()) {
             ans = Math.max(ans, q.peekLast().getValue() - q.peekFirst().getValue() + 1);
-            for (int i = 0, n = q.size(); i < n; ++i) {
-                Pair<TreeNode, Integer> node = q.pollFirst();
-                if (node.getKey().left != null) {
-                    q.offerLast(new Pair<>(node.getKey().left, node.getValue() * 2));
+            for (int n = q.size(); n > 0; --n) {
+                var p = q.pollFirst();
+                root = p.getKey();
+                int i = p.getValue();
+                if (root.left != null) {
+                    q.offer(new Pair<>(root.left, i << 1));
                 }
-                if (node.getKey().right != null) {
-                    q.offerLast(new Pair<>(node.getKey().right, node.getValue() * 2 + 1));
+                if (root.right != null) {
+                    q.offer(new Pair<>(root.right, i << 1 | 1));
                 }
             }
         }
@@ -130,9 +171,49 @@ class Solution {
 }
 ```
 
+```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 {
+    private int ans = 1;
+    private List<Integer> t = new ArrayList<>();
+
+    public int widthOfBinaryTree(TreeNode root) {
+        dfs(root, 0, 1);
+        return ans;
+    }
+
+    private void dfs(TreeNode root, int depth, int i) {
+        if (root == null) {
+            return;
+        }
+        if (t.size() == depth) {
+            t.add(i);
+        } else {
+            ans = Math.max(ans, i - t.get(depth) + 1);
+        }
+        dfs(root.left, depth + 1, i << 1);
+        dfs(root.right, depth + 1, i << 1 | 1);
+    }
+}
+```
+
 ### **C++**
 
-`start * 2` 表示下一层的起点。计算下一层左右子树索引时，减去 `start * 2`，可以防止溢出。
+`i << 1` 表示下一层的起点。计算下一层左右子树索引时，减去 `i << 1`，可以防止溢出。
 
 ```cpp
 /**
@@ -150,23 +231,61 @@ class Solution {
 public:
     int widthOfBinaryTree(TreeNode* root) {
         queue<pair<TreeNode*, int>> q;
-        q.emplace(root, 1);
+        q.push({root, 1});
         int ans = 0;
         while (!q.empty()) {
             ans = max(ans, q.back().second - q.front().second + 1);
-            int start = q.front().second;
-            for (int i = 0, n = q.size(); i < n; ++i) {
-                auto node = q.front();
+            int i = q.front().second;
+            for (int n = q.size(); n; --n) {
+                auto p = q.front();
                 q.pop();
-                if (node.first->left != nullptr) q.emplace(node.first->left, node.second * 2 - start * 2);
-                if (node.first->right != nullptr) q.emplace(node.first->right, node.second * 2 + 1 - start * 2);
+                root = p.first;
+                int j = p.second;
+                if (root->left) q.push({root->left, (j << 1) - (i << 1)});
+                if (root->right) q.push({root->right, (j << 1 | 1) - (i << 1)});
             }
         }
         return ans;
     }
 };
 ```
 
+```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<int> t;
+    int ans = 1;
+    using ull = unsigned long long;
+
+    int widthOfBinaryTree(TreeNode* root) {
+        dfs(root, 0, 1);
+        return ans;
+    }
+
+    void dfs(TreeNode* root, int depth, ull i) {
+        if (!root) return;
+        if (t.size() == depth) {
+            t.push_back(i);
+        } else {
+            ans = max(ans, (int) (i - t[depth] + 1));
+        }
+        dfs(root->left, depth + 1, i << 1);
+        dfs(root->right, depth + 1, i << 1 | 1);
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -178,31 +297,68 @@ public:
  *     Right *TreeNode
  * }
  */
-type Node struct {
-	node *TreeNode
-	idx  int
-}
-
 func widthOfBinaryTree(root *TreeNode) int {
-	q := []Node{{root, 1}}
+	q := []pair{{root, 1}}
 	ans := 0
 	for len(q) > 0 {
-		ans = max(ans, q[len(q)-1].idx-q[0].idx+1)
-		n := len(q)
-		for i := 0; i < n; i++ {
-			node := q[0]
+		ans = max(ans, q[len(q)-1].i-q[0].i+1)
+		for n := len(q); n > 0; n-- {
+			p := q[0]
 			q = q[1:]
-			if node.node.Left != nil {
-				q = append(q, Node{node.node.Left, node.idx * 2})
+			root = p.node
+			if root.Left != nil {
+				q = append(q, pair{root.Left, p.i << 1})
 			}
-			if node.node.Right != nil {
-				q = append(q, Node{node.node.Right, node.idx*2 + 1})
+			if root.Right != nil {
+				q = append(q, pair{root.Right, p.i<<1 | 1})
 			}
 		}
 	}
 	return ans
 }
 
+type pair struct {
+	node *TreeNode
+	i    int
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func widthOfBinaryTree(root *TreeNode) int {
+	ans := 1
+	t := []int{}
+	var dfs func(root *TreeNode, depth, i int)
+	dfs = func(root *TreeNode, depth, i int) {
+		if root == nil {
+			return
+		}
+		if len(t) == depth {
+			t = append(t, i)
+		} else {
+			ans = max(ans, i-t[depth]+1)
+		}
+		dfs(root.Left, depth+1, i<<1)
+		dfs(root.Right, depth+1, i<<1|1)
+	}
+	dfs(root, 0, 1)
+	return ans
+}
+
 func max(a, b int) int {
 	if a > b {
 		return a