Question

965. 单值二叉树

English Version

题目描述

如果二叉树每个节点都具有相同的值，那么该二叉树就是_单值_二叉树。

只有给定的树是单值二叉树时，才返回 true；否则返回 false。

 

示例 1：

输入：[1,1,1,1,1,null,1]
输出：true

示例 2：

输入：[2,2,2,5,2]
输出：false

 

提示：

1. 给定树的节点数范围是 [1, 100]。
2. 每个节点的值都是整数，范围为 [0, 99] 。

解法

方法一

# 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 isUnivalTree(self, root: TreeNode) -> bool:
    def dfs(node):
      if node is None:
        return True
      return node.val == root.val and dfs(node.left) and dfs(node.right)

    return dfs(root)

/**
 * 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 isUnivalTree(TreeNode root) {
    return dfs(root, root.val);
  }

  private boolean dfs(TreeNode root, int val) {
    if (root == null) {
      return true;
    }
    return root.val == val && dfs(root.left, val) && dfs(root.right, val);
  }
}

/**
 * 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 isUnivalTree(TreeNode* root) {
    return dfs(root, root->val);
  }

  bool dfs(TreeNode* root, int val) {
    if (!root) return true;
    return root->val == val && dfs(root->left, val) && dfs(root->right, val);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isUnivalTree(root *TreeNode) bool {
  var dfs func(*TreeNode) bool
  dfs = func(node *TreeNode) bool {
    if node == nil {
      return true
    }
    return node.Val == root.Val && dfs(node.Left) && dfs(node.Right)
  }
  return dfs(root)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function isUnivalTree(root: TreeNode | null): boolean {
  const val = root.val;
  const dfs = (root: TreeNode | null) => {
    if (root == null) {
      return true;
    }
    return root.val === val && dfs(root.left) && dfs(root.right);
  };
  return dfs(root.left) && dfs(root.right);
}

// 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 {
  fn dfs(val: i32, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
    if root.is_none() {
      return true;
    }
    let root = root.as_ref().unwrap().borrow();
    root.val == val && Self::dfs(val, &root.left) && Self::dfs(val, &root.right)
  }
  pub fn is_unival_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    let root = root.as_ref().unwrap().borrow();
    Self::dfs(root.val, &root.left) && Self::dfs(root.val, &root.right)
  }
}