feat: add solutions to lc problem: No.0993 (#2324)

No.0993.Cousins in Binary Tree

Author: yanglbme

solution/0900-0999/0993.Cousins in Binary Tree/README.md

@@ -75,21 +75,21 @@
 class Solution:
     def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
         q = deque([(root, None)])
-        d = 0
+        depth = 0
         p1 = p2 = None
-        d1 = d2 = 0
+        d1 = d2 = None
         while q:
             for _ in range(len(q)):
-                node, fa = q.popleft()
+                node, parent = q.popleft()
                 if node.val == x:
-                    p1, d1 = fa, d
-                if node.val == y:
-                    p2, d2 = fa, d
+                    p1, d1 = parent, depth
+                elif node.val == y:
+                    p2, d2 = parent, depth
                 if node.left:
                     q.append((node.left, node))
                 if node.right:
                     q.append((node.right, node))
-            d += 1
+            depth += 1
         return p1 != p2 and d1 == d2
 ```
 
@@ -111,22 +111,20 @@ class Solution:
  */
 class Solution {
     public boolean isCousins(TreeNode root, int x, int y) {
-        TreeNode p1 = null, p2 = null;
-        int d1 = 0, d2 = 0;
         Deque<TreeNode[]> q = new ArrayDeque<>();
         q.offer(new TreeNode[] {root, null});
-        int d = 0;
-        while (!q.isEmpty()) {
+        int d1 = 0, d2 = 0;
+        TreeNode p1 = null, p2 = null;
+        for (int depth = 0; !q.isEmpty(); ++depth) {
             for (int n = q.size(); n > 0; --n) {
-                var p = q.poll();
-                TreeNode node = p[0], fa = p[1];
+                TreeNode[] t = q.poll();
+                TreeNode node = t[0], parent = t[1];
                 if (node.val == x) {
-                    p1 = fa;
-                    d1 = d;
-                }
-                if (node.val == y) {
-                    p2 = fa;
-                    d2 = d;
+                    d1 = depth;
+                    p1 = parent;
+                } else if (node.val == y) {
+                    d2 = depth;
+                    p2 = parent;
                 }
                 if (node.left != null) {
                     q.offer(new TreeNode[] {node.left, node});
@@ -135,7 +133,6 @@ class Solution {
                     q.offer(new TreeNode[] {node.right, node});
                 }
             }
-            ++d;
         }
         return p1 != p2 && d1 == d2;
     }
@@ -157,34 +154,30 @@ class Solution {
 class Solution {
 public:
     bool isCousins(TreeNode* root, int x, int y) {
-        TreeNode* p1 = nullptr;
-        TreeNode* p2 = nullptr;
-        int d1 = 0, d2 = 0;
         queue<pair<TreeNode*, TreeNode*>> q;
-        q.emplace(root, nullptr);
-        int d = 0;
-        while (!q.empty()) {
+        q.push({root, nullptr});
+        int d1 = 0, d2 = 0;
+        TreeNode *p1 = nullptr, *p2 = nullptr;
+        for (int depth = 0; q.size(); ++depth) {
             for (int n = q.size(); n; --n) {
-                auto [node, fa] = q.front();
+                auto [node, parent] = q.front();
                 q.pop();
                 if (node->val == x) {
-                    p1 = fa;
-                    d1 = d;
-                }
-                if (node->val == y) {
-                    p2 = fa;
-                    d2 = d;
+                    d1 = depth;
+                    p1 = parent;
+                } else if (node->val == y) {
+                    d2 = depth;
+                    p2 = parent;
                 }
                 if (node->left) {
-                    q.emplace(node->left, node);
+                    q.push({node->left, node});
                 }
                 if (node->right) {
-                    q.emplace(node->right, node);
+                    q.push({node->right, node});
                 }
             }
-            ++d;
         }
-        return p1 != p2 && d1 == d2;
+        return d1 == d2 && p1 != p2;
     }
 };
 ```
@@ -199,20 +192,18 @@ public:
  * }
  */
 func isCousins(root *TreeNode, x int, y int) bool {
-	type pair struct{ node, fa *TreeNode }
-	q := []pair{pair{root, nil}}
+	type pair struct{ node, parent *TreeNode }
+	var d1, d2 int
 	var p1, p2 *TreeNode
-	var d, d1, d2 int
-	for len(q) > 0 {
+	q := []pair{{root, nil}}
+	for depth := 0; len(q) > 0; depth++ {
 		for n := len(q); n > 0; n-- {
-			p := q[0]
+			node, parent := q[0].node, q[0].parent
 			q = q[1:]
-			node, fa := p.node, p.fa
 			if node.Val == x {
-				p1, d1 = fa, d
-			}
-			if node.Val == y {
-				p2, d2 = fa, d
+				d1, p1 = depth, parent
+			} else if node.Val == y {
+				d2, p2 = depth, parent
 			}
 			if node.Left != nil {
 				q = append(q, pair{node.Left, node})
@@ -221,19 +212,58 @@ func isCousins(root *TreeNode, x int, y int) bool {
 				q = append(q, pair{node.Right, node})
 			}
 		}
-		d++
 	}
-	return p1 != p2 && d1 == d2
+	return d1 == d2 && p1 != p2
+}
+```
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function isCousins(root: TreeNode | null, x: number, y: number): boolean {
+    let [d1, d2] = [0, 0];
+    let [p1, p2] = [null, null];
+    const q: [TreeNode, TreeNode][] = [[root, null]];
+    for (let depth = 0; q.length > 0; ++depth) {
+        const t: [TreeNode, TreeNode][] = [];
+        for (const [node, parent] of q) {
+            if (node.val === x) {
+                [d1, p1] = [depth, parent];
+            } else if (node.val === y) {
+                [d2, p2] = [depth, parent];
+            }
+            if (node.left) {
+                t.push([node.left, node]);
+            }
+            if (node.right) {
+                t.push([node.right, node]);
+            }
+        }
+        q.splice(0, q.length, ...t);
+    }
+    return d1 === d2 && p1 !== p2;
 }
 ```
 
 <!-- tabs:end -->
 
 ### 方法二：DFS
 
-我们设计一个函数 $dfs(root, fa, d)$，表示从根节点 $root$ 出发，其父节点为 $fa$，深度为 $d$，进行深度优先搜索。
+我们设计一个函数 $dfs(root, parent, depth)$，表示从根节点 $root$ 出发，其父节点为 $parent$，深度为 $depth$，进行深度优先搜索。
 
-在函数中，我们首先判断当前节点是否为空，如果为空，则直接返回。如果当前节点的值为 $x$ 或 $y$，则记录该节点的父节点和深度。然后对当前节点的左右子节点分别调用函数 $dfs$，其中父节点为当前节点，深度为当前深度加 $1$。即 $dfs(root.left, root, d + 1)$ 和 $dfs(root.right, root, d + 1)$。
+在函数中，我们首先判断当前节点是否为空，如果为空，则直接返回。如果当前节点的值为 $x$ 或 $y$，则记录该节点的父节点和深度。然后对当前节点的左右子节点分别调用函数 $dfs$，其中父节点为当前节点，深度为当前深度加 $1$。即 $dfs(root.left, root, depth + 1)$ 和 $dfs(root.right, root, depth + 1)$。
 
 当整棵二叉树遍历完毕后，如果 $x$ 和 $y$ 的深度相同且父节点不同，则返回 $true$，否则返回 $false$。
 
@@ -250,19 +280,19 @@ func isCousins(root *TreeNode, x int, y int) bool {
 #         self.right = right
 class Solution:
     def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
-        def dfs(root, fa, d):
+        def dfs(root, parent, depth):
             if root is None:
                 return
             if root.val == x:
-                t[0] = (fa, d)
-            if root.val == y:
-                t[1] = (fa, d)
-            dfs(root.left, root, d + 1)
-            dfs(root.right, root, d + 1)
+                st[0] = (parent, depth)
+            elif root.val == y:
+                st[1] = (parent, depth)
+            dfs(root.left, root, depth + 1)
+            dfs(root.right, root, depth + 1)
 
-        t = [None, None]
+        st = [None, None]
         dfs(root, None, 0)
-        return t[0][0] != t[1][0] and t[0][1] == t[1][1]
+        return st[0][0] != st[1][0] and st[0][1] == st[1][1]
 ```
 
 ```java
@@ -283,8 +313,8 @@ class Solution:
  */
 class Solution {
     private int x, y;
-    private TreeNode p1, p2;
     private int d1, d2;
+    private TreeNode p1, p2;
 
     public boolean isCousins(TreeNode root, int x, int y) {
         this.x = x;
@@ -293,20 +323,19 @@ class Solution {
         return p1 != p2 && d1 == d2;
     }
 
-    private void dfs(TreeNode root, TreeNode p, int d) {
+    private void dfs(TreeNode root, TreeNode parent, int depth) {
         if (root == null) {
             return;
         }
         if (root.val == x) {
-            p1 = p;
-            d1 = d;
-        }
-        if (root.val == y) {
-            p2 = p;
-            d2 = d;
+            d1 = depth;
+            p1 = parent;
+        } else if (root.val == y) {
+            d2 = depth;
+            p2 = parent;
         }
-        dfs(root.left, root, d + 1);
-        dfs(root.right, root, d + 1);
+        dfs(root.left, root, depth + 1);
+        dfs(root.right, root, depth + 1);
     }
 }
 ```
@@ -326,22 +355,22 @@ class Solution {
 class Solution {
 public:
     bool isCousins(TreeNode* root, int x, int y) {
-        TreeNode *p1, *p2;
         int d1, d2;
-        function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* fa, int d) {
+        TreeNode* p1;
+        TreeNode* p2;
+        function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* parent, int depth) {
             if (!root) {
                 return;
             }
             if (root->val == x) {
-                p1 = fa;
-                d1 = d;
+                d1 = depth;
+                p1 = parent;
+            } else if (root->val == y) {
+                d2 = depth;
+                p2 = parent;
             }
-            if (root->val == y) {
-                p2 = fa;
-                d2 = d;
-            }
-            dfs(root->left, root, d + 1);
-            dfs(root->right, root, d + 1);
+            dfs(root->left, root, depth + 1);
+            dfs(root->right, root, depth + 1);
         };
         dfs(root, nullptr, 0);
         return p1 != p2 && d1 == d2;
@@ -359,24 +388,57 @@ public:
  * }
  */
 func isCousins(root *TreeNode, x int, y int) bool {
-	var p1, p2 *TreeNode
 	var d1, d2 int
-	var dfs func(*TreeNode, *TreeNode, int)
-	dfs = func(root *TreeNode, fa *TreeNode, d int) {
+	var p1, p2 *TreeNode
+	var dfs func(root, parent *TreeNode, depth int)
+	dfs = func(root, parent *TreeNode, depth int) {
 		if root == nil {
 			return
 		}
 		if root.Val == x {
-			p1, d1 = fa, d
+			d1, p1 = depth, parent
+		} else if root.Val == y {
+			d2, p2 = depth, parent
 		}
-		if root.Val == y {
-			p2, d2 = fa, d
-		}
-		dfs(root.Left, root, d+1)
-		dfs(root.Right, root, d+1)
+		dfs(root.Left, root, depth+1)
+		dfs(root.Right, root, depth+1)
 	}
 	dfs(root, nil, 0)
-	return p1 != p2 && d1 == d2
+	return d1 == d2 && p1 != p2
+}
+```
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function isCousins(root: TreeNode | null, x: number, y: number): boolean {
+    let [d1, d2, p1, p2] = [0, 0, null, null];
+    const dfs = (root: TreeNode | null, parent: TreeNode | null, depth: number) => {
+        if (!root) {
+            return;
+        }
+        if (root.val === x) {
+            [d1, p1] = [depth, parent];
+        } else if (root.val === y) {
+            [d2, p2] = [depth, parent];
+        }
+        dfs(root.left, root, depth + 1);
+        dfs(root.right, root, depth + 1);
+    };
+    dfs(root, null, 0);
+    return d1 === d2 && p1 !== p2;
 }
 ```