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Problem solution · Python

Count Dominant Nodes in a Binary Tree

Count Dominant Nodes in a Binary Tree: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
52 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

Depth-first search

For Count Dominant Nodes in a Binary Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 52 lines of Python from the credited upstream file count-dominant-nodes-in-a-binary-tree.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeCount Dominant Nodes in a Binary Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(h) # iterative dfsclass Solution(object):    def countDominantNodes(self, root):        """        :type root: Optional[TreeNode]        :rtype: int        """        def iter_dfs():            ret = [0]*2            stk = [(1, (root, ret))]            while stk:                step, args = stk.pop()                if step == 1:                    u, ret = args                    if u is None:                        continue                    ret1 = [0]*2                    ret2 = [0]*2                    stk.append((2, (u, ret1, ret2, ret)))                    stk.append((1, (u.right, ret2)))                    stk.append((1, (u.left, ret1)))                elif step == 2:                    u, ret1, ret2, ret = args                    mx = max(ret1[1], ret2[1], u.val)                    ret[:] = [ret1[0]+ret2[0]+(1 if mx == u.val else 0), mx]            return ret[0]                return iter_dfs()  # Time:  O(n)# Space: O(h)# dfsclass Solution2(object):    def countDominantNodes(self, root):        """        :type root: Optional[TreeNode]        :rtype: int        """        def dfs(u):            if u is None:                return [0]*2            cnt1, mx1 = dfs(u.left)            cnt2, mx2 = dfs(u.right)            mx = max(mx1, mx2, u.val)            return [cnt1+cnt2+(1 if mx == u.val else 0), mx]                return dfs(root)[0]

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