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.
/**
* 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 maxpath;
int maxPathSumInternal(TreeNode *root) {
if(root == NULL) return 0;
int leftmax = max(maxPathSumInternal(root->left), 0);
int rightmax = max(maxPathSumInternal(root->right), 0);
int totalmax = leftmax + rightmax + root->val;
if(totalmax > maxpath)
maxpath = totalmax;
return max(leftmax + root->val, rightmax + root->val);
}
int maxPathSum(TreeNode *root) {
maxpath = 0xa0000000;
maxPathSumInternal(root);
return maxpath;
}
};
本文介绍了一种解决二叉树最大路径和问题的方法。该问题要求找出从任意节点开始到任意节点结束的最大路径和。文章提供了一个递归算法实现,并通过内部函数维护全局变量来跟踪最大路径和。
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