一、
题目：
655. Print Binary Tree
Print a binary tree in an m*n 2D string array following these rules:
- The row number
mshould be equal to the height of the given binary tree. - The column number
nshould always be an odd number. - The root node's value (in string format) should be put in the exactly middle of the first row it can be put. The column and the row where the root node belongs will separate the rest space into two parts (left-bottom part and right-bottom part). You should print the left subtree in the left-bottom part and print the right subtree in the right-bottom part. The left-bottom part and the right-bottom part should have the same size. Even if one subtree is none while the other is not, you don't need to print anything for the none subtree but still need to leave the space as large as that for the other subtree. However, if two subtrees are none, then you don't need to leave space for both of them.
- Each unused space should contain an empty string
"". - Print the subtrees following the same rules.
Example 1:
Input:
1
/
2
Output:
[["", "1", ""],
["2", "", ""]]
Example 2:
Input:
1
/ \
2 3
\
4
Output:
[["", "", "", "1", "", "", ""],
["", "2", "", "", "", "3", ""],
["", "", "4", "", "", "", ""]]
Example 3:
Input:
1
/ \
2 5
/
3
/
4
Output:
[["", "", "", "", "", "", "", "1", "", "", "", "", "", "", ""]
["", "", "", "2", "", "", "", "", "", "", "", "5", "", "", ""]
["", "3", "", "", "", "", "", "", "", "", "", "", "", "", ""]
["4", "", "", "", "", "", "", "", "", "", "", "", "", "", ""]]
Note: The height of binary tree is in the range of [1, 10].
解答：
/**
* 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<string>> printTree(TreeNode* root) {
if(!root)
return vector<vector<string>>();
int h = depthOfTree(root);
int w = (1 << h) - 1;//1<< n； 2 的你次方
// cout << w << endl;
vector<vector<string>> res(h, vector<string>(w, ""));
printBT(res, root, 0, 0, w - 1);
return res;
}
private:
//
void printBT(vector<vector<string>> & res, TreeNode* root, int h, int begin, int end)
{
if(!root)
return;
int mid = begin + (end - begin) / 2;
res[h][mid] = to_string(root->val);
if(root->left)
printBT(res, root->left, h + 1, begin, mid - 1);
if(root->right)
printBT(res, root->right, h + 1, mid + 1, end);
return;
}
//calculate the depth of tree
int depthOfTree(TreeNode * root)
{
if(!root)
return 0;
return 1 + max(depthOfTree(root->left), depthOfTree(root->right));
}
//
};二、
111. Minimum Depth of Binary Tree
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7return its minimum depth = 2.
解答：
/**
* 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 minDepth(TreeNode* root) {
//方法一：递归法
// if(!root)
// return 0;
// if(!root->left && !root->right)
// return 1;//1表示本身的节点
// if(!root->left)
// return minDepth(root->right) + 1;//1表示本身的节点
// if(!root->right)
// return minDepth(root->left) + 1;
// return 1 + min(minDepth(root->left), minDepth(root->right));//1表示本身的节点
//方法二：利用层序遍历的思想 栈
if(!root)
return 0;
queue<TreeNode*> q;
q.push(root);
int level = 0;
while(!q.empty())
{
++level;
int n = q.size();
for(int i = 0; i < n; ++i)
{
TreeNode* p = q.front();
q.pop();
if(p->left == nullptr && p ->right == nullptr)//均满足则是找到了树最小的路径,return level
{
return level;
}
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
}
}
return level;
}
};三、
题目：
104. Maximum Depth of Binary Tree
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7return its depth = 3.
解答：
/**
* 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 maxDepth(TreeNode* root) {
if(!root)
return 0;
return 1 + max(maxDepth(root->left), maxDepth(root->right));
}
};四、
题目：
637. Average of Levels in Binary Tree
Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
Example 1:
Input:
3
/ \
9 20
/ \
15 7
Output: [3, 14.5, 11]
Explanation:
The average value of nodes on level 0 is 3, on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].
Note:
- The range of node's value is in the range of 32-bit signed integer.
解答：
/**
* 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) {
if(!root)
return res;
queue<TreeNode*> q;
q.push(root);
q.push(nullptr);
while(!q.empty())
{
TreeNode* p = q.front();
q.pop();
if(p == nullptr)
{
double cur_average = cur_sum / cnt;
res.push_back(cur_average);
cur_sum = 0;
cnt = 0;
if(q.size() > 0)
q.push(nullptr);
}
else
{
cur_sum += p->val;
++cnt;
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
}
}
return res;
}
private:
vector<double> res;
double cur_sum = 0;
int cnt = 0;
};