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This approach uses recursion to evaluate the binary boolean tree. Starting from the root, recursively evaluate the left and right children of each node. If a node is a leaf, return its boolean value. If a node is non-leaf, apply the boolean operation defined by its value on its children's evaluations (OR for 2, AND for 3).
Time Complexity: O(n) where n is the number of nodes, since we need to visit each node once.
Space Complexity: O(h) where h is the height of the tree, due to the recursion stack.
1public class TreeNode {
2 public int val;
3 public TreeNode left;
4 public TreeNode right;
5 public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
6 this.val = val;
7 this.left = left;
8 this.right = right;
9 }
10}
11
12public class Solution {
13 public bool EvaluateTree(TreeNode root) {
14 if (root.left == null && root.right == null) {
15 return root.val == 1;
16 }
17 bool left = EvaluateTree(root.left);
18 bool right = EvaluateTree(root.right);
19 return root.val == 2 ? left || right : left && right;
20 }
21}C# implementation is built around a similar recursive idea, with base evaluation of leaf node values and recursion to handle non-leaf nodes' logical operations.
This method employs an iterative approach using a stack to emulate the recursive behavior. By performing a depth-first traversal, it uses a stack to track nodes and their processed children, evaluating each in line with the tree's logic operations.
Time Complexity: O(n), as all nodes are processed.
Space Complexity: O(n), given the stack memory employment.
1
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
bool evaluateTree(TreeNode* root) {
std::stack<std::pair<TreeNode*, bool>> s;
s.push({root, false});
while (!s.empty()) {
auto [node, processed] = s.top();
s.pop();
if (!node->left && !node->right) {
node->val = node->val == 1;
continue;
}
if (!processed) {
s.push({node, true});
if (node->right) s.push({node->right, false});
if (node->left) s.push({node->left, false});
} else {
bool left = node->left->val;
bool right = node->right->val;
node->val = node->val == 2 ? (left || right) : (left && right);
}
}
return root->val;
}The C++ example adopts an iterative stack-based approach pairing TreeNode pointers with a 'processed' flag for each entry. Nodes are revisited after evaluating children, applying logical operations before combining results back up the stack.