872. Leaf-Similar Trees

Description

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 = [1,2,3], 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].

Solutions

Solution 1: DFS

We can use Depth-First Search (DFS) to traverse the leaf nodes of the two trees, storing the values of the leaf nodes in two lists $l_1$ and $l_2$ respectively. Finally, we compare whether the two lists are equal.

Time complexity is $O(n)$, and space complexity is $O(n)$. Here, $n$ is the number of nodes in the tree.

Python3JavaC++GoRustJavaScript

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def leafSimilar(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        def dfs(root: Optional[TreeNode], nums: List[int]) -> None:
            if root.left == root.right:
                nums.append(root.val)
                return
            if root.left:
                dfs(root.left, nums)
            if root.right:
                dfs(root.right, nums)

        l1, l2 = [], []
        dfs(root1, l1)
        dfs(root2, l2)
        return l1 == l2

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean leafSimilar(TreeNode root1, TreeNode root2) {
        List<Integer> l1 = new ArrayList<>();
        List<Integer> l2 = new ArrayList<>();
        dfs(root1, l1);
        dfs(root2, l2);
        return l1.equals(l2);
    }

    private void dfs(TreeNode root, List<Integer> nums) {
        if (root.left == root.right) {
            nums.add(root.val);
            return;
        }
        if (root.left != null) {
            dfs(root.left, nums);
        }
        if (root.right != null) {
            dfs(root.right, nums);
        }
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
        vector<int> l1, l2;
        dfs(root1, l1);
        dfs(root2, l2);
        return l1 == l2;
    }

    void dfs(TreeNode* root, vector<int>& nums) {
        if (root->left == root->right) {
            nums.push_back(root->val);
            return;
        }
        if (root->left) {
            dfs(root->left, nums);
        }
        if (root->right) {
            dfs(root->right, nums);
        }
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
    l1, l2 := []int{}, []int{}
    var dfs func(*TreeNode, *[]int)
    dfs = func(root *TreeNode, nums *[]int) {
        if root.Left == root.Right {
            *nums = append(*nums, root.Val)
            return
        }
        if root.Left != nil {
            dfs(root.Left, nums)
        }
        if root.Right != nil {
            dfs(root.Right, nums)
        }
    }
    dfs(root1, &l1)
    dfs(root2, &l2)
    return reflect.DeepEqual(l1, l2)
}

// 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 leaf_similar(
        root1: Option<Rc<RefCell<TreeNode>>>,
        root2: Option<Rc<RefCell<TreeNode>>>,
    ) -> bool {
        let mut l1 = Vec::new();
        let mut l2 = Vec::new();
        Self::dfs(&root1, &mut l1);
        Self::dfs(&root2, &mut l2);
        l1 == l2
    }

    fn dfs(node: &Option<Rc<RefCell<TreeNode>>>, nums: &mut Vec<i32>) {
        if let Some(n) = node {
            let n = n.borrow();
            if n.left.is_none() && n.right.is_none() {
                nums.push(n.val);
                return;
            }
            if n.left.is_some() {
                Self::dfs(&n.left, nums);
            }
            if n.right.is_some() {
                Self::dfs(&n.right, nums);
            }
        }
    }
}

var leafSimilar = function (root1, root2) {
    const l1 = [];
    const l2 = [];
    const dfs = (root, nums) => {
        if (root.left === root.right) {
            nums.push(root.val);
            return;
        }
        root.left && dfs(root.left, nums);
        root.right && dfs(root.right, nums);
    };
    dfs(root1, l1);
    dfs(root2, l2);
    return l1.join(',') === l2.join(',');
};

872. Leaf-Similar Trees

Description

Solutions

Solution 1: DFS

Comments