leetcode 124. Binary Tree Maximum Path Sum

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.

这题要理解题目意思，否则会提交很多次的。本身这题题目意思也不明确。

来张图说明一下：

对于这个二叉树，最大路径如红线所示，怎么样，意思明白了吧。注意，路线不必经过root。

class Solution {struct TN{int sum;TN*left, *right;TN(int x) :sum(x){left = NULL, right = NULL;}};vector<vector<TN*>>allque;public:~Solution(){if (!allque.empty())for (int i = 0; i < allque[0].size(); i++)delete allque[0][i];}int maxPathSum(TreeNode* root) {if (root == NULL)return 0;if (root->left == NULL&&root->right == NULL)return root->val;vector<TreeNode*>que;que.push_back(root);TN*root1 = new TN(root->val);vector<TN*>que1;que1.push_back(root1);while (!que.empty()){vector<TreeNode*>newque;vector<TN*>newque1;for (int i = 0; i < que.size(); i++){if (que[i]->left != NULL){TN*n = new TN(que[i]->left->val);//n->parent = que1[i];que1[i]->left = n;newque.push_back(que[i]->left);newque1.push_back(n);}if (que[i]->right != NULL){TN*n = new TN(que[i]->right->val);//n->parent = que1[i];que1[i]->right = n;newque.push_back(que[i]->right);newque1.push_back(n);}}allque.push_back(que1);que = newque;que1 = newque1;}int maxsum = -1000000000;int imax = allque.size() - 2;for (int i = imax; i >= 0; i--){for (int j = 0; j < allque[i].size(); j++){int ss = -1000000000;if (allque[i][j]->left != NULL){if (allque[i][j]->left->sum>maxsum)maxsum = allque[i][j]->left->sum;if (allque[i][j]->left->sum>ss)ss = allque[i][j]->left->sum;}if (allque[i][j]->right != NULL){if (allque[i][j]->right->sum > maxsum)maxsum = allque[i][j]->right->sum;if (allque[i][j]->right->sum > ss)ss = allque[i][j]->right->sum;}if (allque[i][j]->left != NULL&&allque[i][j]->right != NULL&&allque[i][j]->right->sum > 0 && allque[i][j]->left->sum > 0){if (allque[i][j]->sum + allque[i][j]->right->sum + allque[i][j]->left->sum > maxsum)maxsum = allque[i][j]->sum + allque[i][j]->right->sum + allque[i][j]->left->sum;}allque[i][j]->sum += (ss > 0 ? ss : 0);if (allque[i][j]->sum > maxsum)maxsum = allque[i][j]->sum;}for (int j = 0; j < allque[i + 1].size(); j++)delete allque[i + 1][j];allque.pop_back();}return maxsum;}};

accept