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22 lines
1.9 KiB
Markdown
22 lines
1.9 KiB
Markdown
9 months ago
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**Objective**: Given a binary tree, find the maximum path sum from one leaf node to another.
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## Idea
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The problem statement suggests that we're dealing with a tree traversal problem where we need to calculate the maximum path sum. The algorithm we're using is a variant of Depth-First Search (DFS).
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Here's the general idea:
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1. **Depth-First Search (DFS)**: We're using a DFS approach to traverse the tree. DFS visits a node's child nodes before visiting its sibling nodes, diving deep into the tree structure before backtracking. This is ideal for our problem because we need to explore all the way from root to leaf nodes to calculate path sums.
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2. **Post-order Traversal**: More specifically, we're using a post-order traversal (left child, right child, root). This allows us to calculate the maximum path sum for the left and right subtrees of each node before calculating the maximum path sum that includes the node itself.
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3. **Maintaining a Global Maximum**: As we calculate the maximum path sum for each node, we compare it with a global maximum value. If it's larger, we update the global maximum. This ensures that we always have the maximum path sum found so far.
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4. **Returning Maximum Branch Sum**: For each node, we return the maximum path sum that includes the node and either its left child or its right child (whichever is larger). This is because a path can't include both the left and right children of a node, as it wouldn't be a valid path (it would form a loop).
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So, we're visiting the tree in a depth-first manner, exploring all paths from the root to the leaf nodes, and keeping track of the maximum path sum found so far.
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## Complexity Analysis
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The time complexity for this algorithm is $O(N)$, where N is the number of nodes in the tree. This is because we visit each node once. The space complexity is $O$(height of the tree), which is the maximum depth of the call stack during the DFS traversal.
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