二叉树 层次遍历 （queue）遍历的变式题（leetcode）

利用队列结构和二叉树，变换求解

116. Populating Next Right Pointers in Each Node

题目地址

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

117. Populating Next Right Pointers in Each Node II

题目地址

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

与 116 题一致

ac代码如下

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
*  int val;
*  TreeLinkNode *left, *right, *next;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if (root == NULL)
return;

queue<TreeLinkNode*> que;
queue<int> queC;

que.push(root);
queC.push(1);

while (!que.empty())
{
TreeLinkNode* top = que.front();
que.pop();
int ceng = queC.front();
queC.pop();

if (que.empty())
{
top->next = NULL;
}
else if (ceng == queC.front())
{
top->next = que.front();
}
else{
top->next = NULL;
}

if (top->left != NULL)
{
que.push(top->left);
queC.push(ceng + 1);
}

if (top->right != NULL)
{
que.push(top->right);
queC.push(ceng + 1);
}
}// end while

}
};

199. Binary Tree Right Side View

题目地址

https://leetcode.com/problems/binary-tree-right-side-view/

题目描述

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:

Given the following binary tree,

1            <---
/   \
2     3         <---
\     \
5     4       <---

You should return [1, 3, 4].

ac代码如下

/**
* 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 Height(TreeNode* root){
if (root == NULL)
return 0;
else if (root->left == NULL && root->right == NULL)
return 1;
else{
return 1 + max(Height(root->left), Height(root->right));
}
}

vector<int> rightSideView(TreeNode* root) {
vector<int> ans;
if (root == NULL)
return ans;

int h = Height(root);

queue<TreeNode*> que;
queue<int> queCeng;
que.push(root);
queCeng.push(1);

int last = 0;
while (!que.empty())
{
TreeNode* top = que.front();
que.pop();
int ceng = queCeng.front();
queCeng.pop();

if (ceng != last)
{
ans.push_back(top->val);
if (ceng == h)
break;
last = ceng;
}
if (top->right != NULL)
{
que.push(top->right);
queCeng.push(ceng + 1);
}
if (top->left != NULL)
{
que.push(top->left);
queCeng.push(ceng + 1);
}
}

return ans;
}
};