Tuesday night. Keep going!!!

amazon.cpp

@@ -470,13 +470,98 @@ public:
             }
             res.push_back(temp);
         }
+        return res;
+    }
+};
+
+
+
 
+Binary Tree Zigzag Level Order Traversal
+//和上一题BFS不同的是，zigzag使用的是2个stack，奇数层用s1，偶数层用s2。
+// 奇数层先看left sub tree,然后right sub tree. 偶数层先看right sub tree, 然后left sub tree
+// https://www.youtube.com/watch?v=YsLko6sSKh8&t=49s
+/**
+ * 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:
+    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
+        vector<vector<int>> res;
+        if(!root) return res;
+        
+        stack<TreeNode*> s1;
+        stack<TreeNode*> s2;
+        
+        s1.push(root);
+        int level = 0;
+        while( !s1.empty()|| !s2.empty()) {
+            if(level%2 == 0){
+                vector<int> temp;
+                while(!s1.empty()){
+                    cout<<"1"<<endl;
+                    TreeNode* curr = s1.top();
+                    s1.pop();
+                    temp.push_back(curr->val);
+                    if(curr->left) s2.push(curr->left);
+                    if(curr->right) s2.push(curr->right);
+                }
+                res.push_back(temp);
+            }
+            else{
+                vector<int> temp;
+                while(!s2.empty()){
+                    cout<<"2"<<endl;
+                    TreeNode* curr = s2.top();
+                    s2.pop();
+                    temp.push_back(curr->val);
+                    if(curr->right) s1.push(curr->right);
+                    if(curr->left) s1.push(curr->left);
+                }
+                res.push_back(temp);
+            }
+            level++;
+            cout<<"3"<<endl;
+        }
         return res;
+
     }
 };
 
 
+Binary Tree Maximum Path Sum
+// not fully understood
+// 大体意思为，因为可以不包含root，所以随时要考虑是否应该用 new_path，来代替包含root的old path
+// 要随时和0比较是因为有可能出现左右分支都小于0的情况 (避免sum有可能被加上了较小的那个负数情况)
+// return的内容是为recursion服务的，传回的值为parent加上一个较大的noded情况
 
+class Solution {
+    int sum = INT_MIN;
+public:
+    int maxPathSum(TreeNode* root) {
+        max_gain(root);
+        return sum;
+    }
+    
+    int max_gain(TreeNode* parent){
+        if(!parent) return 0;
+        
+        int left  = max(max_gain(parent->left), 0);
+        int right = max(max_gain(parent->right), 0);
+        
+        int new_path = left + right + parent->val;
+        
+        sum = max(sum, new_path);
+        
+        return max(parent->val + max(left, right), 0);   
+    } 
+};