[124]. Binary Tree Maximum Path Sum, [543]. Diameter of Binary Tree

[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 must contain at least one node and does not need to go through the root.

For example:Given the below binary tree,       1      / \     2   3Return 6.

这题基本的做法还是可以分成，最大路径出现在左子树，右子树，以及通过根节点三种情况讨论，复杂度为O(nlogn).复杂度理解参照快排。
代码：

class Solution {public:    int maxPathSum(TreeNode* root) {        if (root == NULL) return INT_MIN;        if (root->left == NULL && root->right == NULL) return root->val;        int maxleft = maxPathSum(root->left);        int maxright = maxPathSum(root->right);        int maxltor = root->val;        int maxleftconn = maxconn(root->left, 0);        int maxrightconn = maxconn(root->right, 0);        return max(max(maxleft, maxright), maxltor + maxleftconn + maxrightconn);    }    int maxconn(TreeNode* root, int prevSum) {        if (root == NULL) return prevSum;        if (root->left == NULL && root->right == NULL) return max(prevSum, prevSum + root->val);        int maxleftconn = maxconn(root->left, root->val);        int maxrightconn = maxconn(root->right, root->val);        return max(prevSum, prevSum + max(maxleftconn, maxrightconn));    }};

方法二：
这题其实可以将左右子树中的判断和通过根节点的判断融为一体，减少不必要的递归。
可以记住这种很常用的处理手段。因为实际上所有的判断归根结底都是发生在寻找左边最大路径，右边最大路径，不论是左右相加还是左右取较大，都是基于同样的操作。除了递归，我们可以维持一个maxsum的数来记录所有可能的路径中的最大和。
代码：

class Solution {public:    int maxPathSum(TreeNode* root) {        int maxsum = INT_MIN;        DFS(root, maxsum);        return maxsum;    }    int DFS(TreeNode* root, int& maxsum) {        if (root == NULL) return 0;        int left = max(0, DFS(root->left, maxsum));        int right = max(0, DFS(root->right, maxsum));        maxsum = max(maxsum, root->val + left + right);        return max(root->val, root->val + max(left, right));    }};

复杂度为O(n).
同样的操作可以用在另一题上：
543. Diameter of Binary Tree

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

Example:Given a binary tree           1         / \        2   3       / \           4   5    Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

class Solution {public:    int diameterOfBinaryTree(TreeNode* root) {        int diameter = 0;        DFS(root, diameter);        return diameter;    }    int DFS(TreeNode* root, int& diameter) {        if (root == NULL) return 0;        int left = DFS(root->left, diameter);        int right = DFS(root->right, diameter);        diameter = max(diameter, left + right);        return max(left, right) + 1;    }};