从上到下打印二叉树
题目描述
不分行从上到下打印二叉树
利用队列，打印当前节点时，应将该节点的孩子节点放入队列中。每次取队首元素进行打印
分行从上到下打印？
加入分行的标志节点，一个变量表示当前的节点数，另一个表示下层节点数
void PrintTree(BinaryTreeNode *pRoot)
{
if(pRoot == nullptr)
return;
queue<BinaryTreeNode *>nodes;
nodes.push(pRoot);
int nextLevel = 0;
int toBePrint = 1;
while(!nodes.empty())
{
BinaryTreeNode *pNode = nodes.front();
cout<<pNode->val<<" ";
if(pNode->left !=nullptr)
{
nodes.push(pNode->left);
nextLevel ++;
}
if(pNode->right != nullptr)
{
nodes.push(pNode->right);
nextLevel++;
}
nodes.pop();
toBePrint--;
if(toBePrint == 0)
{
cout<<endl;
tiBePrint = nextLevel;
nextLevel = 0;
}
}
}
leetcode
- 二叉树的层次遍历
给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
/**
* 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>> levelOrder(TreeNode* root) {
vector<vector<int>>res;
if(root ==nullptr) return res;
vector<int>nodes;
queue<TreeNode *>record;
record.push(root);
int currentlevel = 1;
int nextlevel = 0;
while(!record.empty())
{
TreeNode *Node = record.front();
nodes.push_back(Node->val);
if(Node->left!=nullptr)
{
record.push(Node->left);
nextlevel ++;
}
if(Node->right !=nullptr)
{
record.push(Node->right);
nextlevel++;
}
record.pop();
currentlevel--;
if(currentlevel == 0)
{
currentlevel = nextlevel;
nextlevel = 0;
res.push_back(nodes);
nodes.clear();
}
}
return res;
}
};
执行用时 : 12 ms, 在Binary Tree Level Order Traversal的C++提交中击败了95.17% 的用户
内存消耗 : 13.9 MB, 在Binary Tree Level Order Traversal的C++提交中击败了50.53% 的用户
leetcode 二叉树的层次遍历 II
- 二叉树的层次遍历 II
给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）
只需要把上面代码改一句res的插入位置即可
/**
* 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>> levelOrderBottom(TreeNode* root) {
vector<vector<int>>res;
if(root ==nullptr) return res;
vector<int>nodes;
queue<TreeNode *>record;
record.push(root);
int currentlevel = 1;
int nextlevel = 0;
while(!record.empty())
{
TreeNode *Node = record.front();
nodes.push_back(Node->val);
if(Node->left!=nullptr)
{
record.push(Node->left);
nextlevel ++;
}
if(Node->right !=nullptr)
{
record.push(Node->right);
nextlevel++;
}
record.pop();
currentlevel--;
if(currentlevel == 0)
{
currentlevel = nextlevel;
nextlevel = 0;
res.insert(res.begin(), nodes);
nodes.clear();
}
}
return res;
}
};
执行用时 : 20 ms, 在Binary Tree Level Order Traversal II的C++提交中击败了64.92% 的用户
内存消耗 : 14.1 MB, 在Binary Tree Level Order Traversal II的C++提交中击败了30.31% 的用户
leeched 二叉树的层平均值
- 二叉树的层平均值
给定一个非空二叉树, 返回一个由每层节点平均值组成的数组.
同样，修改条件即可
/**
* 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<double> averageOfLevels(TreeNode* root) {
vector<double>res;
if(root ==nullptr) return res;
double sum = 0.0;
queue<TreeNode *>record;
record.push(root);
double avg = 0.0;
int currentlevel = 1;
int count = 1;
int nextlevel = 0;
while(!record.empty())
{
TreeNode *Node = record.front();
sum += Node->val;
if(Node->left!=nullptr)
{
record.push(Node->left);
nextlevel ++;
}
if(Node->right !=nullptr)
{
record.push(Node->right);
nextlevel++;
}
record.pop();
currentlevel--;
if(currentlevel == 0)
{
currentlevel = nextlevel;
nextlevel = 0;
avg = sum / count;
res.push_back(avg);
avg = 0.0;
sum =0.0;
count = currentlevel;
}
}
return res;
}
};
