
Problem Statement
In the given problem, you are provided with the root of a binary tree. Every node in this binary tree contains a value between 0 and 25 inclusive, representing the lowercase English alphabets from 'a' to 'z' (where 0 corresponds to 'a', 1 to 'b', and so on up to 25 which corresponds to 'z'). The task is to determine the lexicographically smallest string that originates at a leaf node and concludes at the root node. A key detail to remember is that a string is considered lexicographically smaller if it is a shorter prefix of another string. A leaf node in this context is defined as a node with no children.
Examples
Example 1
Input:
root = [0,1,2,3,4,3,4]
Output:
"dba"
Example 2
Input:
root = [25,1,3,1,3,0,2]
Output:
"adz"
Example 3
Input:
root = [2,2,1,null,1,0,null,0]
Output:
"abc"
Constraints
- The number of nodes in the tree is in the range
[1, 8500]
. 0 <= Node.val <= 25
Approach and Intuition
The problem can be approached by conducting a depth-first search (DFS) from the root to all the leaves. Given that we are required to find the smallest lexicographical string starting from any leaf to the root, here is a step-by-step method:
Depth-first Search (DFS):
- Initiate a recursive DFS function from the root.
- The DFS will carry the current path as it explores the tree.
- If a leaf node is reached (a node with no children), calculate the string from the leaf to the root by reversing the path compiled during the traversal.
String Comparison:
- Maintain a variable to store the smallest lexicographical string encountered during the DFS.
- Every time a new leaf is reached and a new path string is formed, compare it lexically with the stored string.
- Update the smallest string variable if the newly generated string is smaller.
Backtracking:
- Ensure that once you backtrack in your DFS, the current node data is removed or ignored as you return up the tree. This prevents incorrect path formation.
Edge-cases:
- Handle cases where there's a single node, which is uniquely the smallest string.
- Also, handle cases where there might be multiple paths yielding the same smallest string; ensure consistency by maintaining approach in how paths are compared.
By following through with this strategy, every leaf-to-root path is assessed, guaranteeing that the lexicographically smallest string is correctly identified by the completion of the DFS search. Given the constraints, the approach should run efficiently within the allowable limits.
Solutions
- C++
class Solution {
public:
string findSmallestLeafPath(TreeNode* root) {
string smallest = "";
queue<pair<TreeNode*, string>> workQueue;
workQueue.push({root, string(1, root->val + 'a')});
while (!workQueue.empty()) {
auto[currNode, path] = workQueue.front();
workQueue.pop();
if (!currNode->left && !currNode->right) {
if (smallest == "") {
smallest = path;
} else {
smallest = min(smallest, path);
}
}
if (currNode->left) {
workQueue.push({currNode->left, char(currNode->left->val + 'a') + path});
}
if (currNode->right) {
workQueue.push({currNode->right, char(currNode->right->val + 'a') + path});
}
}
return smallest;
}
};
To solve the problem of finding the smallest string starting from any leaf in a binary tree, where each node represents a letter, the given C++ code illustrates an efficient approach:
- Start by defining a function within a class named
Solution
that will process a given binary tree and return the smallest lexicographical string from the leaves back to the root. - Use a
queue
to perform a breadth-first search (BFS). This queue stores a pair consisting of the current tree node and the string formed by the path to that node. - Initialize the queue with the root node. Convert the value of the node to the corresponding character and form the initial string.
- Enter a loop that continues until the queue is empty. Inside the loop:
- Dequeue the front pair, which includes the current node and the path string.
- Check if the current node is a leaf (no left or right child):
- For leaf nodes, compare the path string with the previously found smallest string. Update the smallest string if the new one is lexicographically smaller.
- For non-leaf nodes:
- If there is a left child, enqueue the left child and append its character to the current path string.
- Similarly, enqueue the right child if it exists, again updating the path string.
- After the loop finishes, the variable
smallest
holds the smallest string formed from leaf to root. Return this string.
This approach ensures you systematically explore all paths from the root to the leaves, updating the smallest string as paths are checked. Through BFS, every node's value contributes to forming potential paths, efficiently comparing and finding the lexicographically smallest string possible from leaf to root.
- Java
class Solution {
public String minimumLeafPath(TreeNode root) {
String minPath = "";
Queue<Pair<TreeNode, String>> traversalQueue = new LinkedList<>();
traversalQueue.add(new Pair<>(root, String.valueOf((char)(root.val + 'a'))));
while (!traversalQueue.isEmpty()) {
Pair<TreeNode, String> treeNodePair = traversalQueue.poll();
TreeNode currentNode = treeNodePair.getKey();
String pathSoFar = treeNodePair.getValue();
if (currentNode.left == null && currentNode.right == null) {
if (minPath.isEmpty()) {
minPath = pathSoFar;
} else {
minPath = pathSoFar.compareTo(minPath) < 0 ? pathSoFar : minPath;
}
}
if (currentNode.left != null) {
traversalQueue.add(new Pair<>(currentNode.left, (char)(currentNode.left.val + 'a') + pathSoFar));
}
if (currentNode.right != null) {
traversalQueue.add(new Pair<>(currentNode.right, (char)(currentNode.right.val + 'a') + pathSoFar));
}
}
return minPath;
}
}
In the provided Java solution, the goal is to find the smallest lexicographical string from the leaf to the root in a binary tree. Each node in the tree contains an integer value that corresponds to a lowercase letter ('a' to 'z', given by 0 to 25). The solution employs a breadth-first search (BFS) using a queue to explore all paths from root to the tree's leaves.
- Initialize an empty string
minPath
to store the smallest path discovered. - Use a
Queue
to facilitate BFS, where each element is aPair
consisting of aTreeNode
and the current path string formed from the root to this node. - The root of the tree is added to the queue initially, with its path string being just its corresponding character.
- A loop runs as long as there are nodes to process in the queue:
- Dequeue the next node and path.
- If the dequeued node is a leaf (i.e., it has no children):
- If
minPath
is still empty or the new path is lexicographically smaller, updateminPath
.
- If
- For each existing child of the current node, construct the new path by adding the child's character to the current path and enqueue the child and this new path.
- Continue this process until all paths leading to leaves are evaluated.
- Return the smallest path
minPath
found, which will be the lexicographically smallest string from any leaf to the root.
This approach efficiently determines the desired path by continuously updating the smallest path found during the BFS traversal.
- Python
class Solution:
def findSmallestString(self, root: Optional[TreeNode]) -> str:
least_val = ""
queue = deque()
queue.append([root, chr(root.val + ord('a'))])
while queue:
node, str_val = queue.popleft()
if not node.left and not node.right:
least_val = min(least_val, str_val) if least_val else str_val
if node.left:
queue.append([node.left, chr(node.left.val + ord('a')) + str_val])
if node.right:
queue.append([node.right, chr(node.right.val + ord('a')) + str_val])
return least_val
The provided Python code defines a method findSmallestString
within the Solution
class aimed at finding the lexicographically smallest string from a given binary tree, where each node's value represents a character (using the English lowercase letters). The method utilizes a breadth-first search strategy, transforming each node's integer value into its corresponding character and building strings as it traverses from the root to the leaves.
Understand the key steps of the code:
- Initialize an empty string
least_val
to store the smallest string. - Use a deque named
queue
to hold nodes along with their corresponding string values as the tree is traversed. - Begin by appending the root node and its character representation to the queue.
- Process each node in the queue:
- Extract the node and its string value.
- If the current node is a leaf (no left or right children):
- Compare the current string with
least_val
to keep track of the smallest string.
- Compare the current string with
- If the node has a left child, append it to the queue, prepending the left child’s character to the current string value.
- If the node has a right child, similarly append it to the queue but prepend the right child's character.
- Finally, return the smallest string found,
least_val
.
This approach ensures that by the end of the traversal, the smallest lexicographical string formed from the root to any leaf in the binary tree is identified and returned.
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