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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.
解题思路，递归
a
b c
curmax = max (a+b, a, a+c) //计算当前节点单边最大值，
如果a+b 最大，那就是说，把左子树包含进来，有利可图
如果a+c 最大，把右子树包含进来，有利可图
如果 a，最大，那就到a为止，最好了。
gmax = max(gmax, max(curmax, a+b+c))
这个是看要不要抛弃前面的结果.
void maxpathsumhelper(TreeNode* root, int& currentsum, int& maxsum) { if(root == NULL)return; int leftsum = 0; int rightsum = 0; maxpathsumhelper(root->left,leftsum, maxsum); maxpathsumhelper(root->right,rightsum, maxsum); currentsum = max(root->val, max(root->val + leftsum, root->val + rightsum)); //this value will passedback maxsum = max(maxsum, max(currentsum, leftsum + rightsum + root->val)); } int maxPathSum(TreeNode *root) { if(root == NULL) return 0; int csum = INT_MIN; int maxsum = INT_MIN; maxpathsumhelper(root, csum, maxsum); return maxsum; }
Binary Tree Maximum Path Sum 解题注意
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原文地址：http://www.cnblogs.com/oceancloud/p/4181531.html
