199. Binary Tree Right Side View (Medium)

最新推荐文章于 2023-11-15 16:06:43 发布

Niko_Ke

最新推荐文章于 2023-11-15 16:06:43 发布

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本文介绍了一个中等难度的LeetCode题目：二叉树右视图。该题要求从二叉树的顶部到底部返回所有可见节点的值。文章提供了一个使用广度优先搜索的C++解决方案，通过队列实现逐层遍历，最终获取每层最右侧节点的值。

Leetcode 199. Binary Tree Right Side View (Medium) (cpp)

Tag: Tree, Depth-first Search, Breadth-first Search

Difficulty: Medium

/*

199. Binary Tree Right Side View (Medium)

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,
1            <---
/   \
2     3         <---
\     \
5     4       <---
You should return [1, 3, 4].

*/
class Solution {
public:
	vector<int> rightSideView(TreeNode* root) {
		if (root == NULL) {
			return{};
		}
		vector<int> res;
		queue<TreeNode*> q;
		q.push(root);
		while (!q.empty()) {
			int _size = q.size();
			for (int i = 0; i < _size; i++) {
				TreeNode* node = q.front();
				q.pop();
				if (node->left != NULL) {
					q.push(node->left);
				}
				if (node->right != NULL) {
					q.push(node->right);
				}
				if (i == _size - 1) {
					res.push_back(node->val);
				}
			}
		}
		return res;
	}
};