26. Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

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

For example: Given the below binary tree,

       1
      / \
     2   3

 

Return 6.

思想： 后序遍历。注意路径的连通： 结点不为空时要返回  max( max(leftV, rightV)+rootV, rootV);

/**
 * 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) {
        getMaxSum(root);
        return _maxPathSum;
    }
protected:
    int getMaxSum(TreeNode *root) {
        if(root == NULL) return Min_Val;
        int left = getMaxSum(root->left);
        int right = getMaxSum(root->right);
        return findMax(left, right, root->val);
    }
    int findMax(int left, int right, int rootV) {
        int PathSum = max(max(left, right)+rootV, rootV);
        _maxPathSum = max(_maxPathSum, max(PathSum, rootV + left + right));
        return PathSum;
    }
private:
    enum{ Min_Val = -1000};
    int _maxPathSum = Min_Val;
};