104. The maximum depth of a binary tree – LeetCode
Every function must run until return before it ends. + 1 directly at the return level means that the depth of the recursion at this level must be calculated.
if you want
if (root->left==nullptr & amp; & amp; root ->right == nullptr)
{
return 1;
}
In this form, you still need to determine whether root is null, so it is not necessary.
This conclusion can serve as the basis for the following questions
110. Balanced Binary Tree – LeetCode
The solution to this problem given in the Silver Challenge is through postorder traversal, that is, from bottom to top
111. Minimum depth of binary tree – LeetCode
@Claude
class Solution {
public:
int minDepth(TreeNode* root) {
if(root==nullptr){
return 0;
}
int min_Depth = INT_MAX;
if(root->left!=nullptr){
min_Depth= min(minDepth(root->left),min_Depth);
}
if(root->right!=nullptr){
min_Depth= min(minDepth(root->right),min_Depth);
}
return min_Depth + 1;
}
};What’s wrong with this code?
Claude
Apply 1 day ago
Analyzing this code, the main problems are as follows:
1. If the root node has only a left subtree or a right subtree, minDepth should return the depth of the left subtree/right subtree. But this code will only correctly return minDepth when neither the left nor right subtree is empty.
2. When the left and right subtrees of the root node are empty, according to the definition of minDepth, 1 should be returned, indicating that the root node is the minimum depth. But this code returns INT_MAX, which is wrong.
3. When calculating the minDepth of the left and right subtrees, only the smaller value is compared with the current minDepth, without considering the situation when the left/right subtree is empty.
4. The boundary case where the root node is empty is not properly considered and 0 should be returned.
5. It is not appropriate to initialize the variable minDepth to INT_MAX, which is more suitable for initializing the maximum depth. Initializing the minimum depth to INT_MAX may cause incorrect results to be returned.
6. It is more standard to name the function parameter root and not rootNode.
7. It is more concise to write the return value directly as return minDepth + 1.
8. Lack of comments and no description of the function.
9. Consider using Math.min instead of min to calculate the minimum value.
10. Function parameter checks should be added to prevent null pointers from being passed in.
In summary, the code has obvious logical flaws, and it is necessary to reconsider how to correctly implement the minimum depth of the calculation tree. The coding style also needs to be improved, with necessary comments added. (edited)
Claude
Apply 1 day ago
Please note: Claude is not skilled at solving math problems.
See the Claude documentation for more information.
xionggavin
1 day ago
@Claude
int minDepth(TreeNode* root) {
if(root==nullptr){
return 0;
}
if(root->left == nullptr & amp; & amp; root->right == nullptr){
return 1;
}
int min_Depth = INT_MAX;
if(root->left!=nullptr){
min_Depth= min(minDepth(root->left),min_Depth);
}
if(root->right!=nullptr){
min_Depth= min(minDepth(root->right),min_Depth);
}
return min_Depth + 1;
}
};In this code if(root->left == nullptr & amp; & amp; root->right == nullptr){
return 1;
}What is the function of this statement? Can it be deleted?
Claude
Apply 1 day ago
What this code does is:
When both the left subtree and the right subtree of the root node root are empty, the minimum depth of 1 is returned directly.
This judgment statement is very important and cannot be deleted.
Because the definition of minimum depth is: the number of nodes on the shortest path from the root node to the nearest leaf node.
If the left and right subtrees of the root node are empty, then the root node itself is a leaf node, and the minimum depth is 1.
If this judgment statement is deleted, when the left and right subtrees of the root node are empty, the minimum depth of the left and right subtrees will continue to be calculated recursively:
– When the left subtree is empty, minDepth(root->left) returns 0
– When the right subtree is empty, minDepth(root->right) returns 0
Finally, min_Depth takes the minimum value of two 0s, 0, and returns it by adding 1, so the minimum depth of 1 will be returned incorrectly.
So the purpose of this sentence is to deal with an important boundary condition:
When the root node is a leaf node, 1 is returned directly.
This avoids the recursive calculation of the depth of the empty subtree and optimizes the code logic.
Therefore, it is very necessary logic and must not be deleted. Deletion will cause the algorithm to fail. (edited)
Claude
Apply 1 day ago
Please note: Claude is not skilled at solving math problems.
See the Claude documentation for more information.
xionggavin
1 day ago
@Claude
What if we ignore the case where there is only the root node?
Claude
Apply 1 day ago
If you ignore the situation where there is only the root node, you delete this code:
if(root->left == nullptr & amp; & amp; root->right == nullptr){
return 1;
}
Two problems will arise:
1. When the tree has only the root node, the minimum depth calculation will be wrong.
If the left and right subtrees of the root node are empty, then the minimum depth should be 1. But if you delete this sentence and directly enter the subsequent logic, INT_MAX will be returned.
This is inconsistent with the definition of minimum depth.
2. The code is not logically rigorous and incomplete.
Consider this sentence from the code:
int min_Depth = INT_MAX;
The purpose is to initialize a maximum value and use it to obtain the minimum value when calculating the minimum depth later.
But if only the root node is not considered, then min_Depth will always remain INT_MAX.
This actually destroys the integrity of the code logic.
Even for more complex trees, the final calculated minimum depth may still be correct.
But this kind of “right” result achieved by “wrong” a boundary condition is very fragile and unstable.
Therefore, for the sake of the rigor and robustness of the code logic, it is still necessary to consider the boundary situation of only the root node.
In short, ignoring this boundary condition will make the code logic incomplete and imprecise, causing hidden dangers. Therefore, this statement still needs to be retained and cannot be ignored. (edited)
xionggavin
1 day ago
@Claude
The question limits the situation where there is only the root node.
xionggavin
1 day ago
@Claude
The question limits the situation where there is only the root node.
Claude
Apply 1 day ago
It turns out that the question has restricted the situation where there is only the root node. In this case, you can consider deleting this code:
if(root->left == nullptr & amp; & amp; root->right == nullptr){
return 1;
}
If it is determined that there will not be only the root node, deleting this code will not affect the correctness.
Analyze it:
1. Due to the problem limit, there will not be a tree with only the root node.
2. Then the root node either has no child nodes or has at least one non-empty child node.
3. If the root node has no child nodes, based on the judgment of the first line of code, 0 will be returned directly.
4. If there is at least one child node, it will enter the subsequent logic to calculate the minimum depth of the left and right subtrees.
5. Finally, return the minimum depth of the left and right subtrees + 1 as the minimum depth of the root node.
It can be seen that after adding restrictions to the question, the function of this code can be replaced by the first line of code:
if(root == nullptr) return 0;
Therefore, this judgment statement can be safely deleted without affecting the correctness of the algorithm.
This is in line with a principle of programming: remove unnecessary code to make the logic clearer and more concise.
To sum up, if the question adds restrictions to ensure that there will not be a tree with only the root node, you can delete this judgment statement to make the code more concise without affecting the correctness. Otherwise, this statement still needs to be retained. (edited)
xionggavin
1 day ago
@Claude
But why does this code still only return INT_MAX directly?
xionggavin
1 day ago
@Claude
@Claude
But why does this code still only return INT_MAX directly?
Claude
Apply 1 day ago
It turns out that the question has restricted the situation where there is only the root node. In this case, you can consider deleting this code:
cpp
if(root->left == nullptr & amp; & amp; root->right == nullptr){
return 1;
}
If it is determined that there will not be only the root node, deleting this code will not affect the correctness.
Analyze it:
1. Due to the problem limit, there will not be a tree with only the root node.
2. Then the root node either has no child nodes or has at least one non-empty child node.
3. If the root node has no child nodes, based on the judgment of the first line of code, 0 will be returned directly.
4. If there is at least one child node, it will enter the subsequent logic to calculate the minimum depth of the left and right subtrees.
5. Finally, return the minimum depth of the left and right subtrees + 1 as the minimum depth of the root node.
It can be seen that after adding restrictions to the question, the function of this code can be replaced by the first line of code:
cpp
if(root == nullptr) return 0;
Therefore, this judgment statement can be safely deleted without affecting the correctness of the algorithm.
This is in line with a principle of programming: remove unnecessary code to make the logic clearer and more concise.
To sum up, if the question adds restrictions to ensure that there will not be a tree with only the root node, you can delete this judgment statement to make the code more concise without affecting the correctness. Otherwise, this statement still needs to be retained. (edited)
Claude
Apply 1 day ago
Your question is very good, let me analyze it:
Even if this judgment statement is deleted:
if(root->left == nullptr & amp; & amp; root->right == nullptr){
return 1;
}
But why can the code still only return INT_MAX?
The main reasons are:
It is inappropriate to initialize minDepth to INT_MAX.
cpp
int minDepth = INT_MAX;
This will cause minDepth to remain at the maximum value and cannot be updated to the minimum depth of the left and right subtrees.
minDepth should be initialized to a number greater than zero, for example:
cpp
int minDepth = 1;
Or directly initialize it to the smaller value of the minimum depth of the left and right subtrees:
cpp
int minDepth = min(minDepth(root->left), minDepth(root->right));
In addition, when calculating the minimum depth of the left and right subtrees, it is also necessary to consider the situation where the left and right subtrees are empty to avoid recursive errors.
So the correct code implementation should be:
cpp
int minDepth(TreeNode* root) {
if (root == nullptr) return 0;
int leftDepth = root->left ? minDepth(root->left) : 0;
int rightDepth = root->right ? minDepth(root->right) : 0;
return 1 + min(leftDepth, rightDepth);
}
Only in this way can the minimum depth from the root node to the leaves be correctly obtained.
To summarize, you need to properly initialize the value of minDepth and handle the situation where the subtree is empty to get the correct solution. (edited)
xionggavin
1 day ago
@Claude
class Solution {
public:
int minDepth(TreeNode *root) {
if (root == nullptr) {
return 0;
}
queue
que.emplace(root, 1);
while (!que.empty()) {
TreeNode *node = que.front().first;
int depth = que.front().second;
que.pop();
if (node->left == nullptr & amp; & amp; node->right == nullptr) {
return depth;
}
if (node->left != nullptr) {
que.emplace(node->left, depth + 1);
}
if (node->right != nullptr) {
que.emplace(node->right, depth + 1);
}
}
return 0;
}
};In this code
Can you explain in detail while (!que.empty()) {
TreeNode *node = que.front().first;
int depth = que.front().second;
que.pop();
if (node->left == nullptr & amp; & amp; node->right == nullptr) {
return depth;
}
if (node->left != nullptr) {
que.emplace(node->left, depth + 1);
}
if (node->right != nullptr) {
que.emplace(node->right, depth + 1);
}
}The function and operation logic of each line of statements below are not
xionggavin
1 day ago
@Claude
@Claude
class Solution {
public:
int minDepth(TreeNode *root) {
if (root == nullptr) {
return 0;
}
queue
que.emplace(root, 1);
while (!que.empty()) {
TreeNode *node = que.front().first;
int depth = que.front().second;
que.pop();
if (node->left == nullptr & amp; & amp; node->right == nullptr) {
return depth;
}
if (node->left != nullptr) {
que.emplace(node->left, depth + 1);
}
if (node->right != nullptr) {
que.emplace(node->right, depth + 1);
}
}
return 0;
}
};In this code
Can you explain in detail while (!que.empty()) {
TreeNode *node = que.front().first;
int depth = que.front().second;
que.pop();
if (node->left == nullptr & amp; & amp; node->right == nullptr) {
return depth;
}
if (node->left != nullptr) {
que.emplace(node->left, depth + 1);
}
if (node->right != nullptr) {
que.emplace(node->right, depth + 1);
}
}The function and operation logic of each line of statements below are not
Claude
Apply 1 day ago
Okay, let me explain the logic of this while loop code in detail:
1. while (!que.empty())
This is a loop condition. When the queue que is not empty, the loop continues.
2. TreeNode *node = que.front().first;
Take a node from the head of the queue and record it as node.
3. int depth = que.front().second;
Take out the depth of this node and record it as depth.
(Because the queue stores pairs, first is the node, and second is the depth)
4.que.pop();
Pop the removed node from the queue.
5. if (node->left == nullptr & amp; & amp; node->right == nullptr) {
return depth;
}
Check whether the node is a leaf node (both left and right children are empty), if so, directly return the depth of the node as the minimum depth result.
6. if (node->left != nullptr) {
que.emplace(node->left, depth + 1);
}
If the node has a left child, enqueue the left child node with depth + 1.
7. if (node->right != nullptr) {
que.emplace(node->right, depth + 1);
}
If the node has a right child, enqueue the right child node with depth + 1.
8. return 0;
If the queue is empty and no leaf nodes are encountered, it means that this is an empty tree and the minimum depth is 0.
In summary, this code uses the idea of BFS sequential traversal to calculate the shortest distance from the root node to the leaf node, that is, the minimum depth of the tree.
The key is to use the queue to record the node and its corresponding depth information. Each time, the head element of the queue is taken out to determine whether it is a leaf. If it is, the current depth is returned directly. It is very subtle. (edited)
xionggavin
1 day ago
@Claude
I have two questions about the above explanation
1. Why (because the queue stores pairs, first is the node, and second is the depth)?
2. What specific basic syntax is involved in que.emplace? Can you explain it in detail?
Claude
Apply 1 day ago
Let me explain the two questions you raised in detail:
1. Why are pairs stored in the queue, with first being the node and second being the depth?
This is the idea of using the C++ pair type to store two pieces of information in the queue at the same time.
When calculating the minimum depth of a tree, it is necessary to traverse all nodes of the tree and record the depth of each node.
Pair can bind each node to the corresponding depth and put it into the queue as a whole.
In this way, when the queue is queued, the depth of the node can be directly obtained, avoiding the complexity of separately storing and maintaining a depth variable.
It is a good design idea to combine related data into a data type (pair in this case).
2. What specific basic syntax is involved in que.emplace? Can you explain it in detail?
que.emplace is an important method of C++ STL queues. Its function is to construct an element directly in the queue.
The specific syntax is as follows:
que.emplace(args);
args are the parameters to construct the element. Unlike queue.push, emplace avoids the overhead of copying or moving elements.
For example here:
que.emplace(node->left, depth + 1);
A pair will be constructed directly in the queue, the first is node->left, and the second is depth + 1.
It avoids the two-step operation of first constructing the pair and then pushing the pair into the queue.
So emplace can insert new elements into the queue more efficiently.
In summary, using pair and emplace, the node and depth information of the tree can be efficiently managed at the same time, achieving elegant minimum depth calculation. (edited)
xionggavin
1 day ago
@Claude
Can you tell me more about 2. TreeNode *node = que.front().first;
Take a node from the head of the queue and record it as node.
3. int depth = que.front().second;
Take out the depth of this node and record it as depth.
(Because the queue stores pairs, first is the node, and second is the depth) How is the depth of this node achieved?
Claude
Apply 1 day ago
Regarding your question about removing nodes and depth, let me explain in detail:
2. TreeNode *node = que.front().first;
Take a node from the head of the queue and record it as node.
This code uses the first element access syntax of pair que.front().first to obtain the first value of the head element, which is the tree node.
3. int depth = que.front().second;
Take out the depth of this node and record it as depth.
This code uses the second element access syntax of pair que.front().second to obtain the second value of the head element, which is the depth of the tree node.
The reason why the two values in the pair can be accessed directly through first and second is because the queue stores data of the pair type.
The definition of pair is:
cpp
pair
In this way, we can directly access the two data through p.first and p.second.
The elements in the queue are of this pair type, so you can directly use que.front().first and que.front().second to get the two values of the queue head element.
in conclusion:
The queue stores data of type pair
Through the first and second access syntax of pair, the tree node and corresponding depth in each element can be directly obtained.
This avoids maintaining a depth variable yourself, making it more concise and efficient to use.
Hope this helps you understand the implementation mechanism of taking out nodes and depth! Any follow-up questions are welcome. (edited)
559. Maximum depth of N-ary tree – LeetCode
xionggavin
1 day ago
@Claude
class Solution {
public:
int maxDepth(Node* root) {
if (root == nullptr) {
return 0;
}
int maxChildDepth = 0;
vector
for (auto child : children) {
int childDepth = maxDepth(child);
maxChildDepth = max(maxChildDepth, childDepth);
}
return maxChildDepth + 1;
}
In this code, vector
for (auto child : children) {
int childDepth = maxDepth(child);
maxChildDepth = max(maxChildDepth, childDepth);
}
return maxChildDepth + 1;
}What does each of the following statements mean? Please explain yx in detail
Claude
Apply 1 day ago
Okay, let me explain this code in detail:
1. vector
This line of code defines a vector to store all child nodes of the current node root. root->children represents the list of child nodes of the current node.
2. for (auto child : children)
This is the range for loop syntax in C++, which is used to traverse each element in the children vector, that is, each child node. The auto keyword will automatically deduce that the type of child is Node*.
3. int childDepth = maxDepth(child);
Recursively call the maxDepth function to calculate the maximum depth of the subtree rooted at the current child, recorded as childDepth.
4. maxChildDepth = max(maxChildDepth, childDepth);
Use the max function to compare the maximum depth childDepth of the current subtree with the maximum depth maxChildDepth of the previously traversed subtree, and assign the maximum value to maxChildDepth.
5. return maxChildDepth + 1;
Return the maximum depth maxChildDepth in all subtrees plus 1, where 1 represents the height of the current node root.
To sum up, this code recursively compares the maximum depth of the subtree rooted at each child node, finds the maximum value, and adds 1 to the maximum depth of the entire tree with the current node root as the root.
It uses the idea of recursive traversal and comparison of maximum values, which is very subtle.
Please let me know if anything is unclear and I’ll be happy to explain it in detail. (edited)
xionggavin
1 day ago
@Claude
2. for (auto child : children)
This is the range for loop syntax in C++, which is used to traverse each element in the children vector, that is, each child node. The auto keyword will automatically deduce that the type of child is Node*. Can you explain in detail the basic syntax involved?
Claude
Apply 1 day ago
You asked about the syntax of C++ range for loop, let me explain it in detail:
The range-based for loop (range-based for loop) was introduced in C++11, and the syntax is as follows:
cpp
for (decltype(auto) element : range) {
// code block
}
This loop will traverse the range specified by range, and assign the next element in the range to the element variable each iteration.
Specifically, let’s look at several key points in the forEach loop:
1. Range – This must be a traversable range, usually an array, container, etc., which determines the iteration range of the loop.
2. decltype(auto) – is a type derivation syntax, used to automatically deduce the type of element, which is equivalent to auto.
3. element – represents the element variable of the traversal iteration, the type is automatically deduced by the compiler.
4. : – Separator, associates element and range.
So in the above code:
cpp
for (auto child : children)
It is to traverse the children vector, and each iteration child will automatically obtain the type of the element (Node*) in children.
The range for loop automatically handles the iteration logic, is simple to use, and is safer and more efficient than the ordinary for loop.
To summarize, the range for loop can implement iteration over a range with very concise code, which is a very useful feature of C++11.
Hope this helps you understand the syntax and advantages of ranged for loops in C++! Please let me know if you have any other questions.