Use a post-order DFS that returns the height of each subtree while updating a global maximum diameter. At each node, the longest path through it is the sum of the heights of its left and right subtrees; update the global max and return 1 + max(left, right).
Pro tip: Clarify that the diameter may or may not pass through the root, and mention that the O(n) solution is optimal because every node must be visited at least once.
Confirm that the diameter is the number of edges (or nodes) on the longest path between any two nodes, and that the path may or may not pass through the root. Ask if the tree is binary and if edge weights are uniform.
Define a helper function that returns the height of a subtree (longest downward path from that node) and updates a global variable for the maximum diameter seen so far.
For a given node, compute the heights of its left and right subtrees. The longest path through this node is left_height + right_height; update the global maximum if this sum is larger.
Return 1 + max(left_height, right_height) to the parent, representing the longest downward path from the current node.
State that the algorithm visits each node once, so time complexity is O(n) and space complexity is O(h) for the recursion stack, where h is the tree height.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than I'd like to admit.
Model the problem as a shortest path between the two islands, where the cost of traversing a cell is 1 if it's water (0) and 0 if it's land (1). Use multi-source BFS from one island to compute the minimum number of water cells to flip to reach the other island, or run BFS from both islands simultaneously and track the minimum sum of distances when frontiers meet.
Pro tip: Clarify that flipping a 0 to 1 connects it to adjacent land, so the answer is the minimum number of water cells on any path between the islands, not the Manhattan distance. Mention that you can optimize by only exploring water cells and treating land as cost 0, or by using 0-1 BFS.
Scan the matrix to find the starting cell of each island (e.g., using DFS/BFS to label connected components).
Decide between multi-source BFS from one island or simultaneous BFS from both. Multi-source BFS from one island is simpler; simultaneous BFS can be more efficient.
Use a queue to explore cells, tracking the number of water cells flipped so far. For 0-1 BFS, use a deque and push water cells to the back and land cells to the front.
When a cell from the other island is reached, return the current cost. If using simultaneous BFS, track the minimum sum of distances when frontiers meet.
State that time and space complexity are O(m*n) where m and n are matrix dimensions, as each cell is visited at most once.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.