leetcode学习笔记4

Symmetric Tree

Given a binarytree, check whether it is a mirror of itself (ie, symmetric around its center).
For example,this binary tree is symmetric:

1
/\
2 2
/ \/ \
3 44 3

But thefollowing is not:

1
/\
2 2
\ \
3 3

class Solution {

public:

bool isSymmetric(TreeNode* root) {

if(!root) return true;

return help(root->left,root->right);

}

private:

bool help(TreeNode* left,TreeNode* right){

if(left==NULL&&right==NULL) return true;

if(left==NULL||right==NULL) return false;

if(left->val!=right->val) return false;

return(help(left->right,right->left)&&help(left->left,right->right));

}

};

注：重点！！要注意返回值，if(left->val==right->val)return true; else return false;是错误的！！易犯错误！！

Balanced Binary Tree

Given a binary tree,determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as abinary tree in which the depth of the two subtrees of every node never differ by more than 1.

class Solution {

public:

bool isBalanced(TreeNode* root) {

if(root==NULL) return true;

int l=maxdep(root->left);

int r=maxdep(root->right);

if(l-r<-1||l-r>1) return false;

else returnisBalanced(root->left)&&isBalanced(root->right);

}

private:

int maxdep(TreeNode* root){

if(root==NULL) return 0;

int l=maxdep(root->left);

int r=maxdep(root->right);

return (max(l,r)+1);

}

};

注：递归的典型题

Implement Queue using Stacks

Implement thefollowing operations of a queue using stacks.

push(x) -- Push element x to the back of queue.
pop() -- Removes the element from in front of queue.

peek() -- Get the front element.
empty() -- Return whether the queue is empty.

class Queue {

public:

// Push element x to the back of queue.

stack<int> s1,s2;

void transfer(){

int x;

if(s2.empty())

while(!s1.empty()){

x=s1.top();

s1.pop();

s2.push(x);

}

}

void push(int x) {

s1.push(x);

}

// Removes the element from in front of queue.

void pop(void) {

transfer();

s2.pop();

}

// Get the front element.

int peek(void) {

int x;

transfer();

x=s2.top();

return x;

}

// Return whether the queue is empty.

bool empty(void) {

if(s1.empty()&&s2.empty()) return true;

else return false;

}

};

注：考察栈和队列的基本操作！！！

Merge Two Sorted Lists

Merge two sortedlinked lists and return it as a new list. The new list should be made bysplicing together the nodes of the first two lists.
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class Solution {

public:

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {

ListNode *head=new ListNode(3);

ListNode *p=head;

while(l1&&l2){

if(l1->val<l2->val){

p->next=l1;

p=l1;

l1=l1->next;

}

else{

p->next=l2;

p=l2;

l2=l2->next;

}

}

while(l1) {

p->next=l1;

p=l1;

l1=l1->next;}

while(l2)

{p->next=l2;

p=l2;

l2=l2->next;}

head=head->next;

return head;

}

};

注:leetcode上的链表，头结点往往带有值，创立一个新的节点，使问题简化很多