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Problem solution · Python

Tree Diameter

Tree Diameter: a Python solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Breadth-first search
Source
Kamyu LeetCode Solutions
Length
136 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Tree Diameter, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 136 lines of Python from the credited upstream file tree-diameter.py.
  • The implementation visibly relies on sequence storage, ordered lookup, cached states.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeTree Diameter · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(|V| + |E|)# Space: O(|E|) # iterative dfsclass Solution(object):    def treeDiameter(self, edges):        """        :type edges: List[List[int]]        :rtype: int        """        def iter_dfs():            result = 0            stk = [(1, (0, -1, [0]))]            while stk:                step, args = stk.pop()                if step == 1:                    u, p, ret = args                    for v in reversed(adj[u]):                        if v == p:                            continue                        ret2 = [0]                        stk.append((2, (ret2, ret)))                        stk.append((1, (v, u, ret2)))                elif step == 2:                    ret2, ret = args                    result = max(result, ret[0]+(ret2[0]+1))                    ret[0] = max(ret[0], ret2[0]+1)            return result                adj = [[] for _ in range(len(edges)+1)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        return iter_dfs()  # Time:  O(|V| + |E|)# Space: O(|E|)# dfsclass Solution2(object):    def treeDiameter(self, edges):        """        :type edges: List[List[int]]        :rtype: int        """        def dfs(u, p):            mx = 0            for v in adj[u]:                if v == p:                    continue                curr = dfs(v, u)                result[0] = max(result[0], mx+(curr+1))                mx = max(mx, curr+1)            return mx                    adj = [[] for _ in range(len(edges)+1)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        result = [0]        dfs(0, -1)        return result[0]  # Time:  O(|V| + |E|)# Space: O(|E|)# bfs, tree dpclass Solution3(object):    def treeDiameter(self, edges):        """        :type edges: List[List[int]]        :rtype: int        """        def bfs():            result = 0            dp = [0]*len(adj)            degree = map(len, adj)            q = [u for u in xrange(len(degree)) if degree[u] == 1]            while q:                new_q = []                for u in q:                    if degree[u] == 0:                        continue                    degree[u] -= 1                    for v in adj[u]:                        if degree[v] == 0:                            continue                        result = max(result, dp[v]+(dp[u]+1))                        dp[v] = max(dp[v], (dp[u]+1))                        degree[v] -= 1                        if degree[v] == 1:                            new_q.append(v)                q = new_q            return result                adj = [[] for _ in range(len(edges)+1)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        return bfs()  # Time:  O(|V| + |E|)# Space: O(|E|)# bfsclass Solution4(object):    def treeDiameter(self, edges):        """        :type edges: List[List[int]]        :rtype: int        """        def bfs(root):            d = new_root = -1            lookup = [False]*len(adj)            lookup[root] = True            q = [root]            while q:                d, new_root = d+1, q[0]                new_q = []                for u in q:                    for v in adj[u]:                        if lookup[v]:                            continue                        lookup[v] = True                        new_q.append(v)                q = new_q            return d, new_root                adj = [[] for _ in range(len(edges)+1)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        _, root = bfs(0)        d, _ = bfs(root)        return d

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