Binary Tree Maximum Path Sum

Description

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

Example
Given the below binary tree:
  1
 / \
2   3
return 6.

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: An integer
     */
    int maxPathSum(TreeNode *root) {
        // write your code here
        if (!root) {
            return 0;
        }
        pair<int, int> result = helper(root);
        return result.second;
    }
private:
    // <int,int> root2any , any2any
    // pair<int, int> pernode;
    pair<int, int> helper(TreeNode* root){
        if (!root) {
            return make_pair(INT_MIN, INT_MIN);
        }
        pair<int, int> left = helper(root->left);
        pair<int, int> right = helper(root->right);
        // update two kinds of path
        // root will be every recursion's root
        int root2any = max(0, max(left.first, right.first)) + root->val;
        int any2any = max(left.second, right.second);
        // find max any to any
        // max any to any,  left any to its root + right any to its root + root
        any2any = max(any2any, max(0, left.first) + max(0, right.first) + root->val);
        return make_pair(root2any, any2any);
    }
};

Similar problems

x

Tags

x