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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

• Java
• C++
• Python
• Go
• RenderScript

• /**
   * 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 = dfs(root1);
          List<Integer> l2 = dfs(root2);
          return l1.equals(l2);
      }
  
      private List<Integer> dfs(TreeNode root) {
          if (root == null) {
              return new ArrayList<>();
          }
          List<Integer> ans = dfs(root.left);
          ans.addAll(dfs(root.right));
          if (ans.isEmpty()) {
              ans.add(root.val);
          }
          return ans;
      }
  }
  

• /**
   * 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) {
          return dfs(root1) == dfs(root2);
      }
  
      vector<int> dfs(TreeNode* root) {
          if (!root) return {};
          auto ans = dfs(root->left);
          auto right = dfs(root->right);
          ans.insert(ans.end(), right.begin(), right.end());
          if (ans.empty()) ans.push_back(root->val);
          return ans;
      }
  };
  

• # 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):
              if root is None:
                  return []
              ans = dfs(root.left) + dfs(root.right)
              return ans or [root.val]
  
          return dfs(root1) == dfs(root2)
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
  	var dfs func(*TreeNode) []int
  	dfs = func(root *TreeNode) []int {
  		if root == nil {
  			return []int{}
  		}
  		ans := dfs(root.Left)
  		ans = append(ans, dfs(root.Right)...)
  		if len(ans) == 0 {
  			ans = append(ans, root.Val)
  		}
  		return ans
  	}
  	return reflect.DeepEqual(dfs(root1), dfs(root2))
  }
  

• // 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;
  impl Solution {
      #[allow(dead_code)]
      pub fn leaf_similar(
          root1: Option<Rc<RefCell<TreeNode>>>,
          root2: Option<Rc<RefCell<TreeNode>>>
      ) -> bool {
          let mut one_vec: Vec<i32> = Vec::new();
          let mut two_vec: Vec<i32> = Vec::new();
  
          // Initialize the two vector
          Self::traverse(&mut one_vec, root1);
          Self::traverse(&mut two_vec, root2);
  
          one_vec == two_vec
      }
  
      #[allow(dead_code)]
      fn traverse(v: &mut Vec<i32>, root: Option<Rc<RefCell<TreeNode>>>) {
          if root.is_none() {
              return;
          }
          if Self::is_leaf_node(&root) {
              v.push(root.as_ref().unwrap().borrow().val);
          }
          let left = root.as_ref().unwrap().borrow().left.clone();
          let right = root.as_ref().unwrap().borrow().right.clone();
          Self::traverse(v, left);
          Self::traverse(v, right);
      }
  
      #[allow(dead_code)]
      fn is_leaf_node(node: &Option<Rc<RefCell<TreeNode>>>) -> bool {
          node.as_ref().unwrap().borrow().left.is_none() &&
              node.as_ref().unwrap().borrow().right.is_none()
      }
  }
  
  

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