Leetcode Study Day 37

Binary Tree Maximum Path Sum

A path 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

Solution

I found a detaild explanation from Leetcode

The way to think of a solution to this is that when we are looking a path in a tree its unidirectional and cannot retrace back what i mean by that is:
    _
  / 1 \ 
 / / \ \ <-----path that goes like a depth first search without backtracking
/ 2   3 v  

So a way to solve this is that if i am at a node i can choose a left or right subtree but if i choose both this is the only subtree that will contain my maximum

I first set my max_sum to INT_MIN.
I can do either either of the options presented:
1.I can choose to take up the left subtree or drop it.
2.I can either choose to take up the right subtree or drop it.
3.I check for a possibility whether if i were to take both left subtree and right subtree would that beat my current max_sum?
Lets consider
   -10
   / \
  9  20
    /  \
   15   7
I do my postorder traversal with a bit of variation:-

int l=max(max_gain(root->left),0);
int r=max(max_gain(root->right),0);
But why?
This is because I have the option to choose the left or right subtree or whether i will just settle with my root value.

So I do my regular postorder traversal and do the above steps
I hit 9

    9
   / \
NULL  NULL

int l=0,r=0(Base condition)
i store the value of 9+0+0 in a variable
Then check if this is greater than maxsum or not is so i update it.
As my max_sum was INT_MIN it gets updated to 9

Now we explore the right tree of root which reaches 15

    15
   / \
NULL  NULL

int l=0,r=0(Base condition)
i store the value of 9+0+0 in a variable
Then check if this is greater than maxsum or not is so i update it.
As my max_sum was 9 it gets updated to 15

Similarly with 7 but 7 doesnt beat the max_sum so nothing happens.

Now we backtrack 20
here int r=7(as 7>0)
     int l=15(as 15>0)
 now i check whether 20+15+7(considering this subtree to be my maximum)
 as 42>15 max_sum=42
 Now what if we dont consider this subtree?

 Then we choose 20 and maximum of its left or right subtree
 so we send return root->val+max(l,r) to our recursion stack
 so when i reach the root it would be like this
           -10
           /  \    <----I considered 15 and 20 because its along a path and is greater than 20+7
          9    35
  int l=9
      r=35
      check whether 9+35+-10=34 beats max_sum
      34<42 so nothing happens and we return -10+max(9,35)=25 to the caller after which we break out of the helper function and we get max_sum as 42.

Full code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxSum = INT_MIN;
    int calculatePath(TreeNode* root){
        if (!root) return 0;
        int left = max(calculatePath(root -> left), 0);
        int right = max(calculatePath(root -> right), 0);
        maxSum = max(maxSum, root -> val + left +right);
        return root -> val + max(left, right);
    }

    int maxPathSum(TreeNode* root) {
        calculatePath(root);
        return maxSum;
    }
};