Merge branch 'DaleStudy:main' into main

Author: smosco

invert-binary-tree/DaleSeo.rs

@@ -0,0 +1,25 @@
+// 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>>>,
+}
+
+use std::rc::Rc;
+use std::cell::RefCell;
+
+// TC: O(n)
+// SC: O(n)
+impl Solution {
+    pub fn invert_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
+        if let Some(node) = root.as_ref() {
+          let left = node.borrow().left.clone();
+          let right = node.borrow().right.clone();
+
+          node.borrow_mut().left = Solution::invert_tree(right);
+          node.borrow_mut().right = Solution::invert_tree(left);
+        }
+        root
+    }
+}

invert-binary-tree/juhui-jeong.java

@@ -0,0 +1,32 @@
+/*
+ * 시간 복잡도: O(1)
+ * 공간 복잡도: O(1)
+ */
+public class Solution {
+    public TreeNode invertTree(TreeNode root) {
+        if (root == null) {
+            return null;
+        }
+
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.offer(root);
+
+        while (!queue.isEmpty()) {
+            TreeNode cur = queue.poll();
+
+            TreeNode temp = cur.left;
+            cur.left = cur.right;
+            cur.right = temp;
+
+            if (cur.left != null) {
+                queue.offer(cur.left);
+            }
+
+            if (cur.right != null) {
+                queue.offer(cur.right);
+            }
+        }
+
+        return root;
+    }
+}