Left Side View of a Tree
Given a binary tree root, return the leftmost node’s value on each level of the tree.
Constraints
n ≤ 100,000wherenis the number of nodes inroot
https://binarysearch.com/problems/Left-Side-View-of-a-Tree
ExamplesPermalink
Example 1Permalink
Input
- root =
Output
- answer =
[0, 5, 1]
SolutionPermalink
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| /** | |
| * class Tree { | |
| * public: | |
| * int val; | |
| * Tree *left; | |
| * Tree *right; | |
| * }; | |
| */ | |
| vector<int> solve(Tree *root) | |
| { | |
| vector<int> res; | |
| if (!root) | |
| return res; | |
| queue<Tree *> q; | |
| q.push(root); | |
| while (!q.empty()) | |
| { | |
| int levelSize = q.size(); | |
| res.push_back(q.front()->val); | |
| while (levelSize--) | |
| { | |
| Tree *node = q.front(); | |
| q.pop(); | |
| if (node->left) | |
| q.push(node->left); | |
| if (node->right) | |
| q.push(node->right); | |
| } | |
| } | |
| return res; | |
| } |
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