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Boundary of binary tree leetcode 2019 07 09

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okay, let's dive into the leetcode problem "boundary of binary tree" and provide a comprehensive tutorial with code examples.
**problem statement (leetcode 545)**
given a binary tree, return the boundary of it. the boundary is defined as follows:
* the root node is always part of the boundary.
* then the boundary consists of left boundary, leaves, and right boundary in order.
* the left boundary is the set of nodes defined as follows:
* the root node is part of the left boundary.
* if a node has a left child, then the left child is part of the left boundary; otherwise, the right child is part of the left boundary.
* repeat the process until you reach a leaf node or the left boundary has no children.
* the leaves are the nodes that have no children.
* the right boundary is the set of nodes defined as follows:
* if a node has a right child, then the right child is part of the right boundary; otherwise, the left child is part of the right boundary.
* repeat the process until you reach a leaf node or the right boundary has no children.
* the right boundary should be in reverse order.
**constraints:**
* the height of the tree is no more than 1000.
* the total number of nodes is no more than 1000.
**example:**
**conceptual breakdown**
the key to solving this problem is to break it down into smaller, manageable sub-problems:
1. **left boundary:** traverse the left side of the tree, prioritizing the left child whenever available.
2. **leaves:** collect all leaf nodes.
3. **right boundary:** traverse the right side of the tree, prioritizing the right child, and store these nodes in reverse order.
**algorithm**
here's the algorithm we'll use:
1. **initialize `boundary`:** create an empty list to store the boundary nodes.
2. **add the root:** if the root is not `none`, add it to the `boundary`.
3. **left boundary:** find and add the left boundary (excluding leaf nodes).
4. **leaves:** find and add the leaf node ...
#BinaryTree #LeetCode #badvalue
binary tree
boundary traversal
LeetCode
algorithm
tree traversal
depth-first search
DFS
breadth-first search
BFS
left boundary
right boundary
leaf nodes
recursion
problem-solving
coding interview
**problem statement (leetcode 545)**
given a binary tree, return the boundary of it. the boundary is defined as follows:
* the root node is always part of the boundary.
* then the boundary consists of left boundary, leaves, and right boundary in order.
* the left boundary is the set of nodes defined as follows:
* the root node is part of the left boundary.
* if a node has a left child, then the left child is part of the left boundary; otherwise, the right child is part of the left boundary.
* repeat the process until you reach a leaf node or the left boundary has no children.
* the leaves are the nodes that have no children.
* the right boundary is the set of nodes defined as follows:
* if a node has a right child, then the right child is part of the right boundary; otherwise, the left child is part of the right boundary.
* repeat the process until you reach a leaf node or the right boundary has no children.
* the right boundary should be in reverse order.
**constraints:**
* the height of the tree is no more than 1000.
* the total number of nodes is no more than 1000.
**example:**
**conceptual breakdown**
the key to solving this problem is to break it down into smaller, manageable sub-problems:
1. **left boundary:** traverse the left side of the tree, prioritizing the left child whenever available.
2. **leaves:** collect all leaf nodes.
3. **right boundary:** traverse the right side of the tree, prioritizing the right child, and store these nodes in reverse order.
**algorithm**
here's the algorithm we'll use:
1. **initialize `boundary`:** create an empty list to store the boundary nodes.
2. **add the root:** if the root is not `none`, add it to the `boundary`.
3. **left boundary:** find and add the left boundary (excluding leaf nodes).
4. **leaves:** find and add the leaf node ...
#BinaryTree #LeetCode #badvalue
binary tree
boundary traversal
LeetCode
algorithm
tree traversal
depth-first search
DFS
breadth-first search
BFS
left boundary
right boundary
leaf nodes
recursion
problem-solving
coding interview