This approach uses a recursive depth-first search (DFS) to explore every path in the binary tree. We calculate the maximum path sum that a node can contribute to its parent, which is the largest possible sum of the path from this node to any leaf node passing through this node. We keep updating the global maximum path sum during the recursion.
Time Complexity: O(n), where n is the number of nodes in the binary tree. We traverse each node exactly once.
Space Complexity: O(h), where h is the height of the tree. This space is used by the call stack for recursion.
1function TreeNode(val, left, right) {
2 this.val = (val===undefined ? 0 : val);
3 this.left = (left===undefined ? null : left);
4 this.right = (right===undefined ? null : right);
5}
6
7function maxPathSum(root) {
8 let maxSum = -Infinity;
9 function dfs(node) {
10 if (node === null) return 0;
11 let leftMax = Math.max(dfs(node.left), 0);
12 let rightMax = Math.max(dfs(node.right), 0);
13 let currentPathSum = node.val + leftMax + rightMax;
14 maxSum = Math.max(maxSum, currentPathSum);
15 return node.val + Math.max(leftMax, rightMax);
16 }
17 dfs(root);
18 return maxSum;
19}In JavaScript, we define a function maxPathSum that uses a nested function dfs to traverse the tree. The function calculates the maximum path sum for each node as a connection of paths through the children. We maintain a global maximum variable maxSum to capture the largest path sum observed during traversal. Leaf node nullity checks are handled by early returns inside the DFS function. This approach is efficient given the constraints and optimizes path calculations locally with every recursive call.
In this approach, we utilize dynamic programming principles alongside tree traversal to compute the maximum path sum. This method is more about reformulating the tree's recursive structure into a form that allows storing intermediate results to optimize operations across the different subtrees. This method is especially useful for understanding different formulations but often overlaps logically with recursive DFS.
Time Complexity: O(n), covering a single pass over all nodes.
Space Complexity: O(h), necessitated by call stack usage designated by the tree height h.
1class TreeNode {
2 int val;
3 TreeNode left;
4 TreeNode right;
5 TreeNode(int x) { val = x; }
6}
7
8public class Solution {
9 int maxSum = Integer.MIN_VALUE;
10 public int maxPathSum(TreeNode root) {
11 calculateMaxSum(root);
12 return maxSum;
13 }
14 private int calculateMaxSum(TreeNode node) {
15 if (node == null) return 0;
16 int leftGain = Math.max(calculateMaxSum(node.left), 0);
17 int rightGain = Math.max(calculateMaxSum(node.right), 0);
18 int priceNewPath = node.val + leftGain + rightGain;
19 maxSum = Math.max(maxSum, priceNewPath);
20 return node.val + Math.max(leftGain, rightGain);
21 }
22}In this solution, we employ a recursive method calculateMaxSum that computes sub-results by leveraging already seen sub-tree sub-results. By traversing each node once, we update a global max value maxSum with results derived from each node acting as a center of a new path. This method hence avoids recomputation by relying on previously calculated results, drawing dynamic programming parallels.