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题目
You are given the root of a full binary tree with the following properties:
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Leaf nodes have either the value 0 or 1, where 0 represents False and 1 represents True.
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Non-leaf nodes have either the value 2 or 3, where 2 represents the boolean OR and 3 represents the boolean AND.
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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.
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A full binary tree is a binary tree where each node has either 0 or 2 children.
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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.
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思路
仍然是DFS，需要对每个node进行判断：
- 如果node是null，直接返回True
- 如果node的val等于0或者1，直接返回val值
- 如果node的val等于2，需要进行二元操作or
- 如果node的val等于3，需要进行二元操作and
优化思路：
- 题目中标注该二叉树为full binary tree，即完全二叉树
- 当node的左节点为空的时候，则该节点为叶子节点
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代码
python
# 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 not root.left:
return root.val
if root.val == 2:
return self.evaluateTree(root.left) or self.evaluateTree(root.right)
if root.val == 3:
return self.evaluateTree(root.left) and self.evaluateTree(root.right)