Question

2331. Evaluate Boolean Binary Tree

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Description

You are given the root of a full binary tree with the following properties:

• Leaf nodes have either the value 0 or 1, where 0 represents False and 1 represents True.
• Non-leaf nodes have either the value 2 or 3, where 2 represents the boolean OR and 3 represents the boolean AND.

The evaluation of a node is as follows:

• If the node is a leaf node, the evaluation is the value of the node, i.e. True or False.
• Otherwise, evaluate the node's two children and apply the boolean operation of its value with the children's evaluations.

Return_ the boolean result of evaluating the _root_ node._

A full binary tree is a binary tree where each node has either 0 or 2 children.

A leaf node is a node that has zero children.

 

Example 1:

Input: root = [2,1,3,null,null,0,1]
Output: true
Explanation: The above diagram illustrates the evaluation process.
The AND node evaluates to False AND True = False.
The OR node evaluates to True OR False = True.
The root node evaluates to True, so we return true.

Example 2:

Input: root = [0]
Output: false
Explanation: The root node is a leaf node and it evaluates to false, so we return false.

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 3
• Every node has either 0 or 2 children.
• Leaf nodes have a value of 0 or 1.
• Non-leaf nodes have a value of 2 or 3.

Solutions

Solution 1

# 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 evaluateTree(self, root: Optional[TreeNode]) -> bool:
    def dfs(root):
      if root.left is None and root.right is None:
        return bool(root.val)
      l, r = dfs(root.left), dfs(root.right)
      return (l or r) if root.val == 2 else (l and r)

    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 evaluateTree(TreeNode root) {
    return dfs(root);
  }

  private boolean dfs(TreeNode root) {
    if (root.left == null && root.right == null) {
      return root.val == 1;
    }
    boolean l = dfs(root.left), r = dfs(root.right);
    if (root.val == 2) {
      return l || r;
    }
    return l && r;
  }
}

/**
 * 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 evaluateTree(TreeNode* root) {
    return dfs(root);
  }

  bool dfs(TreeNode* root) {
    if (!root->left && !root->right) return root->val;
    bool l = dfs(root->left), r = dfs(root->right);
    if (root->val == 2) return l || r;
    return l && r;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func evaluateTree(root *TreeNode) bool {
  var dfs func(*TreeNode) bool
  dfs = func(root *TreeNode) bool {
    if root.Left == nil && root.Right == nil {
      return root.Val == 1
    }
    l, r := dfs(root.Left), dfs(root.Right)
    if root.Val == 2 {
      return l || r
    }
    return l && r
  }
  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 evaluateTree(root: TreeNode | null): boolean {
  const { val, left, right } = root;
  if (left == null) {
    return val === 1;
  }
  if (val === 2) {
    return evaluateTree(left) || evaluateTree(right);
  }
  return evaluateTree(left) && evaluateTree(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(root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
    let root = root.as_ref().unwrap().as_ref().borrow();
    if root.left.is_none() {
      return root.val == 1;
    }
    if root.val == 2 {
      return Self::dfs(&root.left) || Self::dfs(&root.right);
    }
    Self::dfs(&root.left) && Self::dfs(&root.right)
  }

  pub fn evaluate_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    Self::dfs(&root)
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   struct TreeNode *left;
 *   struct TreeNode *right;
 * };
 */
bool evaluateTree(struct TreeNode* root) {
  if (!root->left) {
    return root->val == 1;
  }
  if (root->val == 2) {
    return evaluateTree(root->left) || evaluateTree(root->right);
  }
  return evaluateTree(root->left) && evaluateTree(root->right);
}

Solution 2

# 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 evaluateTree(self, root: Optional[TreeNode]) -> bool:
    if root.left is None:
      return bool(root.val)
    l = self.evaluateTree(root.left)
    r = self.evaluateTree(root.right)
    return l or r if root.val == 2 else l and r

/**
 * 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 evaluateTree(TreeNode root) {
    if (root.left == null) {
      return root.val == 1;
    }
    boolean l = evaluateTree(root.left);
    boolean r = evaluateTree(root.right);
    return root.val == 2 ? l || r : l && r;
  }
}

/**
 * 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 evaluateTree(TreeNode* root) {
    if (!root->left) {
      return root->val;
    }
    bool l = evaluateTree(root->left);
    bool r = evaluateTree(root->right);
    return root->val == 2 ? l or r : l and r;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func evaluateTree(root *TreeNode) bool {
  if root.Left == nil {
    return root.Val == 1
  }
  l, r := evaluateTree(root.Left), evaluateTree(root.Right)
  if root.Val == 2 {
    return l || r
  }
  return l && r
}