Evaluate Boolean Binary Tree

Problem

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.

Examples

Example 1:

graph TD;
    A(OR) --- B(True) & C(AND);

	C --- D(False) & E(True)

Result in:

graph TD;
    A(OR) --- B(True) & C(False);

Results in:

graph TD;
    A(True)

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.

Solution

Method 1 - Recursive DFS

We can perform postorder DFS.

Video Explanation

Code

Java

class Solution {
    public boolean evaluateTree(TreeNode root) {
	    switch(root.val) {
		    case 0:
			    return false;
			case 1:
				return true;
			case 2: 
				return evaluateTree(root.left) || evaluateTree(root.right);
			default: // case 3
				return evaluateTree(root.left) && evaluateTree(root.right);
				
	    }
    }
}

Complexity

⏰ Time complexity: O(n)
🧺 Space complexity: O(1) (assuming recursive stack is not counted, O(n) otherwise)

Problem#

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Solution#

Method 1 - Recursive DFS#

Video Explanation#

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Java#

Complexity#