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Topic One: Merging two sorted lists

Input two monotonically increasing list, output two list of linked lists, of course, we need to synthesize the linked list to meet the monotone non-reduction rules.

Idea: Use two pointers CUR1,CUR2 to refer to the current smaller and larger nodes in the two linked lists, if cur1->val>cur2->val, then swap the two

/*struct ListNode {int val;    struct ListNode *next; ListNode (int x): Val (x), Next (NULL) {}};*/classSolution { Public: ListNode* Merge (listnode* pHead1, listnode*pHead2) {ListNode* Pre =NULL; ListNode* Next =NULL; if(phead1==null| | phead2==NULL) {            returnPhead1!=null?phead1:phead2; } ListNode* head = Phead1->val<phead2->val?phead1:phead2; ListNode* Cur1 = Head==phead1?phead1:phead2; ListNode* Cur2 = Head==phead1?Phead2:phead1;  while(cur1!=null&&cur2!=NULL) {            if(cur1->val<=cur2->val) {Pre=Cur1; Cur1= cur1->Next; }Else{Next=Cur1; Pre->next =CUR2; Pre=CUR2; Cur1=CUR2; CUR2=Next; }} Pre->next = Cur1!=null?CUR1:CUR2; returnHead; }};

Topic Two: substructure of the tree

Enter two binary trees, a, B, to determine whether a is a sub-structure of a. (PS: We agree that an empty tree is not a sub-structure of any tree)

Idea: Judge B is a sub-structure, so each judgment is to determine a node Suba and b root node can find an equal tree, this equality is conditional, when the node of B traversal to empty, if the Suba is not empty, then it is equal.

So write a function that determines if it is "equal", and then let each node of a get into the function comparison with the root node of B.

/*struct TreeNode {int val;    struct TreeNode *left;    struct TreeNode *right; TreeNode (int x): Val (x), left (null), right (null) {}};*/classSolution { Public:    BOOLIsEqual (treenode* pRoot1, treenode*PRoot2) {        if(Proot2==null)return true ; if(Proot1==null)return false ; if(Proot1->val!=proot2->val)return false ; if(IsEqual (Proot1->left,proot2->left) &&isequal (proot1->right,proot2->right))return true ; return false ; }    BOOLHassubtree (treenode* pRoot1, treenode*PRoot2) {        if(proot2==null| | Proot1==null)return false ; if(IsEqual (Proot1,proot2))return true ; returnHassubtree (proot1->left,proot2) | | Hassubtree (proot1->Right,proot2); }};

Topic 3:2 Fork Tree Mirror

Operates a given two-fork tree and transforms it into a mirror of the source binary tree. Input Description:

Image definition of binary tree: source binary tree         8       /6 x        /     \     5  7 9    Mirror binary tree        8       /        10   6     /\  /     9 7  5
Idea: is simply the right and left child exchange, recursive implementation is good

/*struct TreeNode {int val;    struct TreeNode *left;    struct TreeNode *right; TreeNode (int x): Val (x), left (null), right (null) {}};*/classSolution { Public:    voidMirror (TreeNode *proot) {        if(Proot==null)return ; Mirror (Proot-Left ); Mirror (Proot-Right ); TreeNode*temp = proot->Left ; Proot->left = proot->Right ; Proot->right =temp; }};

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