Problem Description

In this problem, we're given the root of a binary tree. Our goal is to find how many nodes in this tree have values that are equal to the sum of the values of all its descendants. A descendant of a node x is any node that lies in the subtree rooted at x, including all the leaf nodes down the path from node x. If a node has no descendants (i.e., it is a leaf node), the sum of its descendants is considered to be 0. The task is to return the count of such nodes.

To simplify, let's say we have a node with the value 10 and it has two children with values 3 and 7, respectively. The sum of the values of the descendants is 3+7=10, which equals the node's value. This node should be counted as one. We need to do this for every node in the tree.

Flowchart Walkthrough

To analyze the problem LeetCode 1973. Count Nodes Equal to Sum of Descendants using the Flowchart, let's proceed step-by-step through the decision-making process:

Is it a graph?

• Yes: The tree in the problem can be considered as a form of a graph where each node represents a vertex and the connections between nodes represent edges.

Is it a tree?

• Yes: The problem explicitly mentions that the structure is a binary tree.

At this point, since the structure is established to be a tree, we transition directly to employing Depth-First Search (DFS) as suggested by the flowchart. DFS is appropriate because it allows us to explore each node's children and accumulate their values, which directly aligns with the need to sum the descendants’ values at each node to compare with the node's own value.

Conclusion: The flowchart suggests using DFS for this tree-based hierarchical sum comparison problem.

Intuition

The approach uses a classic tree traversal technique known as Depth-First Search (DFS). The DFS allows us to reach the leaf nodes and then go up the tree, summarizing the values as we go along. This particular solution is a post-order traversal, meaning we process a node after we've processed its children.

Here's the intuition behind the approach:

1. Perform a DFS starting from the root.
2. While traversing, calculate the sum of the values of the left and right subtrees individually.
3. After obtaining the sum of the left and right subtree for a node, we check if the sum is equal to the node's value.
4. If so, we increment a counting variable, ans, as we've found a node where the value equals the sum of its descendants.
5. After checking the condition, we return to the parent node the sum of the node's value plus all the values from its descendants.
6. We incrementally build these sums from the child nodes up to the root, performing the checks at each step.
7. The nonlocal keyword used inside the dfs function is a way to modify the ans variable defined outside the nested function's scope.

The traversal ends when all nodes are visited, and the resulting ans variable holds the count of nodes that meet the criteria. This is both an efficient and reliable way to determine the sum of descendants for each node in the tree since with post-order traversal, we visit all the children before the parent, ensuring we have all the necessary information when the sums are compared.

Learn more about Tree, Depth-First Search, Binary Search Tree and Binary Tree patterns.

Solution Approach

The solution leverages a recursive Depth-First Search (DFS) pattern to traverse the binary tree and calculate the sum of the descendants of each node while also keeping track of the count of nodes that fulfill the given criteria.

Here's a detailed breakdown of the implementation:

• Recursive DFS Function (dfs): The dfs function is defined within the equalToDescendants method. This function is a helper function that is called recursively to perform the post-order traversal.

  • It takes a single parameter, the current root of the subtree being processed.
  • Base Case: If we encounter a None node, it signifies that we have reached past a leaf node, and we return 0 since there are no descendants to sum up.
  • Recursive Case: If the node is not None, we recursively call dfs on the left and right children of the current node.
  • We sum up the results of dfs calls on the left and right children, l and r respectively, to get the total sum of the descendants.
  • We compare the sum of the descendant values l + r with the current node's value root.val. If they are equal, we increment the ans counter. This is done using the nonlocal keyword to modify the ans variable that is defined in the outer function's scope.
  • The dfs function returns the sum of the node's value and its descendants' values to its parent node. This is root.val + l + r.

• Recursive Traversal: The recursion stack unwinds, visiting all nodes from the bottom-up (post-order), ensuring each node is checked precisely once for the condition. This helps to optimize the function as the calculated sums for the subtrees are used directly without recalculating them.

• Global Counter (ans): A counter named ans is used to keep track of the number of nodes whose values equal the sum of their descendants' values. It is defined in the same scope as the dfs function, but outside of it, to be accessible and modifiable from within the nested dfs function.

• Finalizing the Count: After initiating the traversal with dfs(root), the function equalToDescendants ultimately returns the value of ans, which is the total number of nodes that satisfy the condition at the end of the traversal.

The time complexity of the solution is O(N) where N is the number of nodes in the tree, as each node is visited exactly once. The space complexity is also O(N), which is the worst-case space used by the call stack during the DFS (this occurs in the case of a skewed tree).

In summary, by using a post-order DFS and a simple counter, we effectively check the sum of descendants for each node against its value and return the total count of such nodes.

Example Walkthrough

Let's illustrate the solution approach with a small binary tree example. Consider the following binary tree:

       10
      /  \
     4    6
    / \
   3   1

We want to find out how many nodes have values equal to the sum of the values of their descendants.

Following the steps given in the solution approach:

1. We start the DFS from the root node which is 10.
2. We traverse to the left child 4 and then to it's left child 3 which is a leaf node.
   • Leaf nodes are base cases, returning 0 since they have no descendants.
3. Back to node 4, we now traverse its right child which is 1 and since it's a leaf node, it also returns 0.
4. Now we're back at 4 with the sum of its descendant being 3 + 0 + 1 + 0 = 4, which equals the node's value.
   • Therefore, we increment our counter ans by 1.
5. We return 4 (node's value) + the sum of its descendants 4, which is 8 to the parent node 10.
6. We proceed to node 10's right child 6, which is a leaf node.
   • Thus, it returns 0 as the sum of the descendants.
7. Now at the root node 10, we take the sum of descendant values we got from the left 8 and right 0 child and compare it with the node 10's value.
   • 8 + 0 is not equal to 10, hence ans remains the same.
8. The function equalToDescendants returns the final ans value which is 1, so there is 1 node with a value equal to the sum of its descendants.

In this example, the node with the value 4 is the only node meeting the criteria.

By following this post-order DFS approach, we calculate the sum of descendant nodes' values for each node and compare it to each node's own value, incrementing our counter when they match and finally returning the count of such nodes.

Solution Implementation

Time and Space Complexity

The given Python code defines a method equalToDescendants which checks each node in a binary tree to see if its value is equal to the sum of the values of its descendant nodes. Let's analyze the time and space complexity of this code:

Time Complexity:

The time complexity of the equalToDescendants method is O(n), where n is the number of nodes in the binary tree. This efficiency results from the fact that the method involves a depth-first search (DFS), visiting each node exactly once. The comparison and addition operations at each node occur in constant time, so the overall time complexity is linear concerning the number of nodes.

Space Complexity:

The space complexity of the code is O(h), where h is the height of the tree. This space is used by the call stack during the execution of the DFS. In the worst case, when the binary tree degrades to a linked list (completely unbalanced), the space complexity will be O(n). In the best case, with a completely balanced tree, the space complexity will be O(log n) because the height of the tree would be logarithmic relative to the number of nodes.

Therefore, the space complexity ranges from O(log n) to O(n) depending on the shape of the tree.

Learn more about how to find time and space complexity quickly using problem constraints.