Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

Total Accepted: 55016 Total Submissions: 246213 Difficulty: Hard

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 does not need to go through the root.

For example:
Given the below binary tree,

       1
      / 
     2   3

Return 6.

 (E) Path Sum (M) Sum Root to Leaf Numbers

 

设最大路径和为sum , 那么sum = max(sum,左子树根出发的最大路径和+根结点的值+右子树根出发的最大路径和);

根据这个式子写成递归式就可以了

有一点需要注意，如果从左右子树根出发的最大路径和为负数，那么将其值置0，因为根的值+从左右子树根出发的最大路径和 一定会小于根的值

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode *root) {
        int maxPath = INT_MIN;
        maxPathWithNode(root,maxPath);
        return maxPath;
    }
    int maxPathWithNode(TreeNode* node,int& maxPath){
        if(!node) return 0;
        int l = max(0,maxPathWithNode(node->left,maxPath));
        int r = max(0,maxPathWithNode(node->right,maxPath));
        maxPath = max(maxPath,l+r+node->val);
       // cout<<"node->val="<<node->val<<" l="<<l<<" r="<<r<<" maxPath="<<maxPath<<endl;
        return max(l,r)+node->val;
    }
};
/*
[-1,-2,-3,4,-5,-6,-7,8,-9,-1,-1,-1,-1,-1,-1,-7,-8]
[-9,-9]
*/