feat: add solutions to lcof problem: No.68 - |

Author: YangFong

lcof/面试题68 - I. 二叉搜索树的最近公共祖先/README.md

@@ -153,6 +153,120 @@ var lowestCommonAncestor = function (root, p, q) {
 };
 ```
 
+### **TypeScript**
+
+递归：
+
+```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 lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: TreeNode | null): TreeNode | null {
+    if (root == null) {
+        return root
+    }
+    if (root.val > p.val && root.val > q.val) {
+        return lowestCommonAncestor(root.left, p, q)
+    }
+    if (root.val < p.val && root.val < q.val) {
+        return lowestCommonAncestor(root.right, p, q)
+    }
+    return root
+};
+```
+
+循环：
+
+```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 lowestCommonAncestor(
+    root: TreeNode | null,
+    p: TreeNode | null,
+    q: TreeNode | null,
+): TreeNode | null {
+    if (root == null) {
+        return root;
+    }
+    while (true) {
+        if (root.val > p.val && root.val > q.val) {
+            root = root.left;
+        } else if (root.val < p.val && root.val < q.val) {
+            root = root.right;
+        } else {
+            return root;
+        }
+    }
+}
+```
+
+### **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::rc::Rc;
+use std::cell::RefCell;
+use std::cmp::Ordering;
+impl Solution {
+    pub fn lowest_common_ancestor(
+        mut root: Option<Rc<RefCell<TreeNode>>>,
+        p: Option<Rc<RefCell<TreeNode>>>,
+        q: Option<Rc<RefCell<TreeNode>>>,
+    ) -> Option<Rc<RefCell<TreeNode>>> {
+        let p = p.unwrap().borrow().val;
+        let q = q.unwrap().borrow().val;
+        loop {
+            let mut cur = root.as_ref().unwrap().borrow().val;
+            match (cur.cmp(&p), cur.cmp(&q)) {
+                (Ordering::Less, Ordering::Less) => root = root.unwrap().borrow().right.clone(),
+                (Ordering::Greater, Ordering::Greater) => {
+                    root = root.unwrap().borrow().left.clone()
+                }
+                (_, _) => break root,
+            }
+        }
+    }
+}
+```
+
 ### **...**
 
 ```

lcof/面试题68 - I. 二叉搜索树的最近公共祖先/Solution.rs

@@ -0,0 +1,41 @@
+// 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::rc::Rc;
+use std::cell::RefCell;
+use std::cmp::Ordering;
+impl Solution {
+    pub fn lowest_common_ancestor(
+        mut root: Option<Rc<RefCell<TreeNode>>>,
+        p: Option<Rc<RefCell<TreeNode>>>,
+        q: Option<Rc<RefCell<TreeNode>>>,
+    ) -> Option<Rc<RefCell<TreeNode>>> {
+        let p = p.unwrap().borrow().val;
+        let q = q.unwrap().borrow().val;
+        loop {
+            let mut cur = root.as_ref().unwrap().borrow().val;
+            match (cur.cmp(&p), cur.cmp(&q)) {
+                (Ordering::Less, Ordering::Less) => root = root.unwrap().borrow().right.clone(),
+                (Ordering::Greater, Ordering::Greater) => {
+                    root = root.unwrap().borrow().left.clone()
+                }
+                (_, _) => break root,
+            }
+        }
+    }
+}

lcof/面试题68 - I. 二叉搜索树的最近公共祖先/Solution.ts

@@ -0,0 +1,31 @@
+/**
+ * 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 lowestCommonAncestor(
+    root: TreeNode | null,
+    p: TreeNode | null,
+    q: TreeNode | null,
+): TreeNode | null {
+    if (root == null) {
+        return root;
+    }
+    while (true) {
+        if (root.val > p.val && root.val > q.val) {
+            root = root.left;
+        } else if (root.val < p.val && root.val < q.val) {
+            root = root.right;
+        } else {
+            return root;
+        }
+    }
+}