LeetCode-in-All
226. Invert Binary Tree
Easy
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Solution
// Definition for a binary tree node.
// 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;
impl Solution {
pub fn invert_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
root.map(|node| {
let node = node.borrow();
let mut new_node = TreeNode::new(node.val);
new_node.left = Self::invert_tree(node.right.clone());
new_node.right = Self::invert_tree(node.left.clone());
Rc::new(RefCell::new(new_node))
})
}
}