問題描述：

Given a non-empty 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.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

源碼：

樹的問題當然離不開遞歸了。找到以每個點為起點的路徑的最大值。時間86%，空間100%。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int result = INT_MIN;
    int change(TreeNode* root){
        if(!root)     return 0;
        int L = max(change(root->left), 0);
        int R = max(change(root->right), 0);
        result = max(result, root->val+L+R);
        return root->val+max(L,R);
    }
    int maxPathSum(TreeNode* root) {
        if(!root)   return 0;
        change(root);
        return result;
    }
};

有遞歸自然就有非遞歸，用一個帶make_pair的棧，模擬兩個遞歸的變量可以，但是我們今天只看看后序遍歷的方法：

時間和空間都和遞歸是一樣的。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        if(!root)   return 0;
        stack<TreeNode*> st;
        int result = INT_MIN;
        st.push(root);
        TreeNode* pre = NULL;
        while(!st.empty()){
            TreeNode* tmp = st.top();
            if((!tmp->left && !tmp->right) || (pre && (pre==tmp->left || pre==tmp->right))){
                st.pop();
                pre = tmp;
                int L, R;
                if(tmp->left)   L = max(tmp->left->val, 0);
                if(tmp->right)  R = max(tmp->right->val, 0);
                if(tmp->left && tmp->right){
                    result = max(result, L + R + tmp->val);
                    tmp->val += max(L, R);
                }
                else if(tmp->left){
                    result = max(result, L + tmp->val);
                    tmp->val += L;
                }
                else if(tmp->right){
                    result = max(result, R + tmp->val);
                    tmp->val += R;
                }
                else{
                    result = max(result, tmp->val);
                }
            }
            else{
                if(tmp->right)  st.push(tmp->right);
                if(tmp->left)   st.push(tmp->left);
            }
        }
        return result;
    }
};