124. Binary Tree Maximum Path Sum (DFS) 20.4.20 Hard

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree 
along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6
Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42


Solution: ----------------------------------- DFS ----------------------- https://www.youtube.com/watch?v=9ZNky1wqNUw
--------------------------------------- O(n) T, O(n) S
# Definition for a binary tree node.
# class TreeNode(object):                               
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def maxPathSum(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        self.res = -float('inf')
        self.dfs(root)                                    
        return self.res
    
    def dfs(self, currnode):                             # For a tree, certainly we need to use recursion(DFS or BFS)
        if not currnode:
            return 0
        leftsum = max(0, self.dfs(currnode.left))        # Classic DFS.
                                                         # IMPORTANT !!!!! We must compare the value with 0 and take the bigger one
                                                         # Because if the leftsum is smaller than 0, we rather not take it into the path
        rightsum = max(0, self.dfs(currnode.right))
        currsum = leftsum + rightsum + currnode.val      # The time complexity is O(n) because we just take every node for consideration
                                                         # The current maximum can take both the left and right substree value (闭环)
        self.res = max(self.res, currsum)
        return currnode.val + max(leftsum, rightsum)     # However, for passing the value to upper recursion,
                                                         # we can only take the larger one of the left or right substree value
                                                         # because the sub-闭环 cannot be a part of a bigger path



Java Version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int maxSum = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        if (root == null) {
            return 0;
        }
        dfs(root);
        return maxSum;
    }
    
    public int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftSum = Math.max(0, dfs(root.left));
        int rightSum = Math.max(0, dfs(root.right));
        int currSum = root.val + leftSum + rightSum;
        maxSum = Math.max(maxSum, currSum);
        return Math.max(root.val + leftSum, root.val + rightSum);
    }
}