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solutions/117. Populating Next Right Pointers in Each Node II.md

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+# [117. Populating Next Right Pointers in Each Node II](https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/)
+
+# 思路
+题目要求将二叉树同一层的节点用指针串起来，要求空间复杂度O(1)。
+这题是[116](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)的升级版，116中的二叉树是满二叉树，比较简单，但思路其实也是类似的，可先参考[116题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/116.%20Populating%20Next%20Right%20Pointers%20in%20Each%20Node.md)。
+## 思路一
+最好想的还是递归: 如果root的左右子树都已经串好了的话，那么我们只需要将左子树每一层的最右边的节点的next指向右子树对应层最左边节点即可。
+
+## 思路二
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    Node* connect(Node* root) {
+        if(root == NULL) return NULL;
+        Node *start_left = connect(root -> left);
+        Node *start_right = connect(root -> right);
+        while(start_left && start_right){
+            Node *last_left = start_left;
+            while(last_left -> next) last_left = last_left -> next;
+            last_left -> next = start_right;
+            
+            while(!start_left->left && 
+                  !start_left->right && 
+                  start_left -> next != start_right) 
+                start_left = start_left -> next;
+            if(start_left -> left) start_left = start_left -> left;
+            else start_left = start_left -> right;
+
+            
+            while(!start_right->left && 
+                  !start_right->right && 
+                  start_right -> next) 
+                start_right = start_right -> next;
+            if(start_right -> left) start_right = start_right -> left;
+            else start_right = start_right -> right;
+            
+        }
+        return root;
+    }
+};
+```
+
+## 思路二