Question

872. 叶子相似的树

English Version

题目描述

请考虑一棵二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。

举个例子，如上图所示，给定一棵叶值序列为 (6, 7, 4, 9, 8) 的树。

如果有两棵二叉树的叶值序列是相同，那么我们就认为它们是 _叶相似 _的。

如果给定的两个根结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。

 

示例 1：

输入：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]
输出：true

示例 2：

输入：root1 = [1,2,3], root2 = [1,3,2]
输出：false

 

提示：

• 给定的两棵树结点数在 [1, 200] 范围内
• 给定的两棵树上的值在 [0, 200] 范围内

解法

方法一：DFS

后序遍历。

# 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.
 * 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.
 * 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()
  }
}