LeetCode [124] Binary Tree Maximum Path Sum

 124. Binary Tree Maximum Path Sum

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

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//C++: 32ms /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int maxPathSum(TreeNode* root) { if(!root) return 0; int ret = INT_MIN; dfs(root, ret); return ret; } int dfs(TreeNode* root, int &ret){ if(!root) return 0; int l = dfs(root->left, ret); int r = dfs(root->right, ret); int len = root->val; if(l>0) len += l; if(r>0) len += r; ret = max(ret, len); return root->val+max(0, max(l, r)); } };

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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { int maxLen = Integer.MIN_VALUE; public int maxPathSum(TreeNode root) { dfs(root); return maxLen; } int dfs(TreeNode node){ if(node==null) return 0; int l = dfs(node.left); int r = dfs(node.right); int len = node.val; if(l>0) len += l; if(r>0) len += r; maxLen = Math.max(maxLen, len); return node.val+Math.max(0, Math.max(l, r)); } }