124. Binary Tree Maximum Path Sum

Given a 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.

For example:
Given the below binary tree,

       1      / \     2   3

Return 6.

My way to solve it is simple:

for each of the node, suppose it is the highest node in the path, compute the maxPath sum

The Code:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    int max = Integer.MIN_VALUE;    HashMap<TreeNode, Integer> hm = new HashMap<>();    public int maxPathSum(TreeNode root) {        helper(root);        return max;    }        void helper(TreeNode root){        if(root == null) return;        int left = maxPathDown(root.left);        int right = maxPathDown(root.right);        int sum = root.val;        if(left > 0) sum += left;        if(right > 0) sum += right;        if(sum > max) max = sum;        helper(root.left);        helper(root.right);    }        int maxPathDown(TreeNode root){        if(root == null) return 0;        if(hm.containsKey(root)) return hm.get(root);        int ans = root.val + Math.max(Math.max(maxPathDown(root.left),maxPathDown(root.right)), 0);        hm.put(root,ans);        return ans;    }}

The helper() method compute and update the path sum for each node as the highest node

The maxPathDown() method compute one path down from a node to the leaf and return the max value of the sum it encountered.

HashMap is used to avoid repeating computation.

I believe this solution runs in O(n) time since every Node is visited 2 times(once in helper and once in maxPathdown)

A smarter way to solve it is to do the update while inside the recursion

public class Solution {    int max = Integer.MIN_VALUE;    HashMap<TreeNode, Integer> hm = new HashMap<>();    public int maxPathSum(TreeNode root) {        maxPathDown(root);        return max;    }        int maxPathDown(TreeNode root){        if(root == null) return 0;        int left = maxPathDown(root.left);        int right = maxPathDown(root.right);        int sum = root.val;        if(left > 0) sum += left;        if(right > 0) sum += right;        if(sum > max) max = sum;        return root.val + Math.max(0,Math.max(left,right));    }}