feat: add solutions to lcof problem: No.28

面试题 28. 对称的二叉树

Author: YangFong

lcof/面试题28. 对称的二叉树/README.md

@@ -190,6 +190,95 @@ public:
 
 ```
 
+### **TypeScript**
+
+```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 isSymmetric(root: TreeNode | null): boolean {
+    if (root == null) {
+        return true;
+    }
+    const dfs = (l: TreeNode | null, r: TreeNode | null) => {
+        if (l == null && r == null) {
+            return true;
+        }
+        if (l == null || r == null) {
+            return false;
+        }
+        return (
+            l.val == r.val &&
+            dfs(l.left, r.right) &&
+            dfs(l.right, r.left) &&
+            true
+        );
+    };
+    return dfs(root.left, root.right);
+}
+```
+
+### **Rust**
+
+```rust
+// Definition for a binary tree node.
+// #[derive(Debug, PartialEq, Eq)]
+// pub struct TreeNode {
+//   pub val: i32,
+//   pub left: Option<Rc<RefCell<TreeNode>>>,
+//   pub right: Option<Rc<RefCell<TreeNode>>>,
+// }
+//
+// impl TreeNode {
+//   #[inline]
+//   pub fn new(val: i32) -> Self {
+//     TreeNode {
+//       val,
+//       left: None,
+//       right: None
+//     }
+//   }
+// }
+use std::cell::RefCell;
+use std::rc::Rc;
+impl Solution {
+    pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
+        match root {
+            None => true,
+            Some(root) => {
+                fn dfs(
+                    l: &Option<Rc<RefCell<TreeNode>>>,
+                    r: &Option<Rc<RefCell<TreeNode>>>,
+                ) -> bool {
+                    if l.is_none() && r.is_none() {
+                        return true;
+                    }
+                    if l.is_none() || r.is_none() {
+                        return false;
+                    }
+                    let l = l.as_ref().unwrap().borrow();
+                    let r = r.as_ref().unwrap().borrow();
+                    l.val == r.val && dfs(&l.left, &r.right) && dfs(&l.right, &r.left) && true
+                }
+                let node = root.borrow();
+                dfs(&node.left, &node.right)
+            }
+        }
+    }
+}
+
+```
 ### **...**
 
 ```

lcof/面试题28. 对称的二叉树/Solution.rs

@@ -0,0 +1,45 @@
+// Definition for a binary tree node.
+// #[derive(Debug, PartialEq, Eq)]
+// pub struct TreeNode {
+//   pub val: i32,
+//   pub left: Option<Rc<RefCell<TreeNode>>>,
+//   pub right: Option<Rc<RefCell<TreeNode>>>,
+// }
+//
+// impl TreeNode {
+//   #[inline]
+//   pub fn new(val: i32) -> Self {
+//     TreeNode {
+//       val,
+//       left: None,
+//       right: None
+//     }
+//   }
+// }
+use std::cell::RefCell;
+use std::rc::Rc;
+impl Solution {
+    pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
+        match root {
+            None => true,
+            Some(root) => {
+                fn dfs(
+                    l: &Option<Rc<RefCell<TreeNode>>>,
+                    r: &Option<Rc<RefCell<TreeNode>>>,
+                ) -> bool {
+                    if l.is_none() && r.is_none() {
+                        return true;
+                    }
+                    if l.is_none() || r.is_none() {
+                        return false;
+                    }
+                    let l = l.as_ref().unwrap().borrow();
+                    let r = r.as_ref().unwrap().borrow();
+                    l.val == r.val && dfs(&l.left, &r.right) && dfs(&l.right, &r.left) && true
+                }
+                let node = root.borrow();
+                dfs(&node.left, &node.right)
+            }
+        }
+    }
+}

lcof/面试题28. 对称的二叉树/Solution.ts

@@ -0,0 +1,34 @@
+/**
+ * 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 isSymmetric(root: TreeNode | null): boolean {
+    if (root == null) {
+        return true;
+    }
+    const dfs = (l: TreeNode | null, r: TreeNode | null) => {
+        if (l == null && r == null) {
+            return true;
+        }
+        if (l == null || r == null) {
+            return false;
+        }
+        return (
+            l.val == r.val &&
+            dfs(l.left, r.right) &&
+            dfs(l.right, r.left) &&
+            true
+        );
+    };
+    return dfs(root.left, root.right);
+}