What if you could find the strongest path in a complex tree without checking every route by hand?
Why DP on Trees Maximum Path Sum in DSA Go?
Imagine you have a family tree drawn on paper, and you want to find the path from one family member to another that has the highest total happiness score. Doing this by checking every possible path manually is like trying to find a needle in a haystack.
Manually checking every path in a tree is slow and confusing because the number of paths grows very fast. You might miss some paths or repeat checking the same parts again and again, making it easy to get tired and make mistakes.
Dynamic Programming on Trees helps by breaking the problem into smaller parts and remembering the best results for each part. This way, you only calculate each part once and combine them to find the maximum path sum efficiently.
func maxPathSumManual(root *Node) int {
// Check all paths by exploring every possible route
// Very slow and complicated
return 0
}func maxPathSumDP(root *Node) int {
if root == nil {
return 0
}
maxSum := math.MinInt32
max := func(a, b int) int {
if a > b {
return a
}
return b
}
var dfs func(*Node) int
dfs = func(node *Node) int {
if node == nil { return 0 }
left := max(0, dfs(node.Left))
right := max(0, dfs(node.Right))
maxSum = max(maxSum, left + right + node.Val)
return max(left, right) + node.Val
}
dfs(root)
return maxSum
}This technique lets you quickly find the best path sum in any tree, even very big ones, without checking every path manually.
Finding the strongest connection path in a social network tree where each connection has a strength score, to recommend the best chain of friends.
Manual path checking is slow and error-prone.
DP on Trees breaks the problem into smaller parts and remembers results.
This leads to fast and reliable maximum path sum calculation.
Practice
Solution
Step 1: Understand the problem definition
The maximum path sum is about finding the path between any two nodes that gives the highest sum of node values.Step 2: Differentiate from other sums
It is not just the sum of all nodes or the longest path length, but the maximum sum along any path.Final Answer:
The largest sum of node values on any path between two nodes in the tree -> Option AQuick Check:
Maximum path sum = largest sum on any path [OK]
- Confusing maximum path sum with longest path length
- Thinking it is sum of all nodes
- Considering only single node values
Solution
Step 1: Identify correct parameter type
The function should accept a pointer to TreeNode to handle nil nodes and avoid copying.Step 2: Identify correct return type
The function returns an integer representing the maximum path sum, so return type int is correct.Final Answer:
func maxPathSum(node *TreeNode) int -> Option DQuick Check:
Pointer input and int output for max sum [OK]
- Passing node by value instead of pointer
- Returning string or error instead of int
- Using wrong parameter type like int
package main
import "fmt"
type TreeNode struct {
Val int
Left, Right *TreeNode
}
func maxPathSumUtil(node *TreeNode, maxSum *int) int {
if node == nil {
return 0
}
left := max(0, maxPathSumUtil(node.Left, maxSum))
right := max(0, maxPathSumUtil(node.Right, maxSum))
*maxSum = max(*maxSum, left + right + node.Val)
return max(left, right) + node.Val
}
func maxPathSum(root *TreeNode) int {
maxSum := -1000000
maxPathSumUtil(root, &maxSum)
return maxSum
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
func main() {
root := &TreeNode{Val: -10}
root.Left = &TreeNode{Val: 9}
root.Right = &TreeNode{Val: 20}
root.Right.Left = &TreeNode{Val: 15}
root.Right.Right = &TreeNode{Val: 7}
fmt.Println(maxPathSum(root))
}Solution
Step 1: Calculate max path sums for subtrees
Left subtree: 9 (updates maxSum to 9). Right subtree: 15 (maxSum=15), 7 (no change), at 20: left=15, right=7, *maxSum = max(15, 15+7+20)=42, returns max(15,7)+20=35.Step 2: Compare and update maxSum
At root (-10): left=max(0,9)=9, right=max(0,35)=35, *maxSum = max(42, 9+35+(-10))=max(42,34)=42.Final Answer:
42 -> Option CQuick Check:
Max path sum = 42 [OK]
- Not ignoring negative sums by using max(0, ...)
- Adding node values incorrectly
- Confusing max path sum with max root-to-leaf sum
func maxPathSumUtil(node *TreeNode, maxSum *int) int {
if node == nil {
return 0
}
left := maxPathSumUtil(node.Left, maxSum)
right := maxPathSumUtil(node.Right, maxSum)
*maxSum = max(*maxSum, left + right + node.Val)
return max(left, right) + node.Val
}Solution
Step 1: Check handling of negative child sums
The function uses left and right sums directly without ignoring negative values, which can reduce the path sum.Step 2: Identify fix by ignoring negatives
To fix, use max(0, left) and max(0, right) to ignore negative sums before adding to node.Val.Final Answer:
It does not ignore negative sums from children before adding -> Option AQuick Check:
Ignore negative child sums for correct max path [OK]
- Not using max(0, child) to ignore negatives
- Returning sum of both children instead of max child
- Not updating maxSum properly
Solution
Step 1: Handle negative child sums carefully
Use max(0, child sums) to ignore negative paths that reduce the total sum.Step 2: Update global maximum path sum correctly
Update the global max with the sum of left + right + current node value to consider paths passing through the node.Final Answer:
Use max(0, child sums) to ignore negative paths and update global max with left + right + node.Val -> Option BQuick Check:
Ignore negatives, update max with left+right+node [OK]
- Summing all nodes including negatives blindly
- Considering only root-to-leaf paths
- Removing negative nodes instead of ignoring sums
Ready to prove it?
Pass the test and this pattern moves unknown
