Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.
For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).
Two binary trees are considered leaf-similar if their leaf value sequence is the same.
Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.
Example 1:
Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8] Output: true
Example 2:
Input: root1 = [1], root2 = [1] Output: true
Example 3:
Input: root1 = [1], root2 = [2] Output: false
Example 4:
Input: root1 = [1,2], root2 = [2,2] Output: true
Example 5:
Input: root1 = [1,2,3], root2 = [1,3,2] Output: false
Constraints:
- The number of nodes in each tree will be in the range
[1, 200]. - Both of the given trees will have values in the range
[0, 200].
struct Solution;
use rustgym_util::*;
trait Leaves {
fn preorder(&self, leaves: &mut Vec<i32>);
}
impl Leaves for TreeLink {
fn preorder(&self, leaves: &mut Vec<i32>) {
if let Some(node) = self {
let node = node.borrow();
let left = &node.left;
let right = &node.right;
if left.is_none() && right.is_none() {
leaves.push(node.val);
} else {
left.preorder(leaves);
right.preorder(leaves);
}
}
}
}
impl Solution {
fn leaf_similar(root1: TreeLink, root2: TreeLink) -> bool {
let mut leaves1: Vec<i32> = vec![];
let mut leaves2: Vec<i32> = vec![];
root1.preorder(&mut leaves1);
root2.preorder(&mut leaves2);
leaves1 == leaves2
}
}
#[test]
fn test() {
let root1 = tree!(
3,
tree!(5, tree!(6), tree!(2, tree!(7), tree!(4))),
tree!(1, tree!(9), tree!(8))
);
let root2 = tree!(
4,
tree!(5, tree!(6), tree!(2, tree!(7), tree!(4))),
tree!(1, tree!(9), tree!(8))
);
assert_eq!(Solution::leaf_similar(root1, root2), true);
}