Binary Tree Maximum Path Sum LeetCode Solution

Minimum Cost to Merge Stones
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Binary Tree Maximum Path Sumpath in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node’s values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

  • The number of nodes in the tree is in the range [1, 3 * 104].
  • -1000 <= Node.val <= 1000

Binary Tree Maximum Path Sum Solutions

Time: O(n)
Space: O(h)

C++

class Solution {
 public:
  int maxPathSum(TreeNode* root) {
    int ans = INT_MIN;
    maxPathSumDownFrom(root, ans);
    return ans;
  }

 private:
  // root->val + 0/1 of its subtrees
  int maxPathSumDownFrom(TreeNode* root, int& ans) {
    if (!root)
      return 0;

    const int l = max(0, maxPathSumDownFrom(root->left, ans));
    const int r = max(0, maxPathSumDownFrom(root->right, ans));
    ans = max(ans, root->val + l + r);
    return root->val + max(l, r);
  }
};

Java

class Solution {
  public int maxPathSum(TreeNode root) {
    maxPathSumDownFrom(root);
    return ans;
  }

  private int ans = Integer.MIN_VALUE;

  // root->val + 0/1 of its subtrees
  private int maxPathSumDownFrom(TreeNode root) {
    if (root == null)
      return 0;

    final int l = Math.max(maxPathSumDownFrom(root.left), 0);
    final int r = Math.max(maxPathSumDownFrom(root.right), 0);
    ans = Math.max(ans, root.val + l + r);
    return root.val + Math.max(l, r);
  }
}

Python


class Solution:
  def maxPathSum(self, root: Optional[TreeNode]) -> int:
    ans = -math.inf

    def maxPathSumDownFrom(root: Optional[TreeNode]) -> int:
      nonlocal ans
      if not root:
        return 0

      l = max(0, maxPathSumDownFrom(root.left))
      r = max(0, maxPathSumDownFrom(root.right))
      ans = max(ans, root.val + l + r)
      return root.val + max(l, r)

    maxPathSumDownFrom(root)
    return ans

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