LeetCode 199. Binary Tree Right Side View

Given the root of 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.

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

实现思路:

谁知道20年的pat就考到这题了呢？姥姥竟然上leetcode抄题目，常规操作了。
采用两种方法实现，bfs+dfs，其中bfs多补充一种重要思路。

AC代码：

bfs方法一

利用计算层数的办法保留每一层的右边第一个

class Solution {
		queue<TreeNode*> q;
		bool vist[110]= {false}; //记录每一层最右边的第一个结点是否访问
		vector<int> ans;
		unordered_map<TreeNode*,TreeNode*> father;
		unordered_map<TreeNode*,int> high;
	public:

		void bfs(TreeNode *root) {
			if(root==nullptr) return;
			q.push(root);
			high[root]=1;
			while(!q.empty()) {
				TreeNode *now=q.front();
				if(!vist[high[now]]) {
					ans.push_back(now->val);
					vist[high[now]]=1;
				}
				q.pop();
				if(now->right) {
					father[now->right]=now;
					high[now->right]=high[now]+1;
					q.push(now->right);
				}
				if(now->left) {
					father[now->left]=now;
					high[now->left]=high[now]+1;
					q.push(now->left);
				}
			}
		}

		vector<int> rightSideView(TreeNode* root) {
			bfs(root);
			return ans;
		}
};

bfs方法二

这里应用到了一个重点的方法，就是size--，层序遍历把一层结点全部出队的做法，在很多题目中都有应用，要掌握。

class Solution {
	public:
		vector<int> rightSideView(TreeNode* root) {
			vector<int> res;
			if (!root) return res;
			queue<TreeNode*> q;
			q.push(root);
			while (q.size()) {
				int size = q.size();
				while (size--) {//先判断进入后马上-1
					auto t = q.front();
					q.pop();
					if (!t) continue;                           //判断空节点
					if (size==0) res.push_back(t->val);   //存入该层最右边一个节点
					if (t->left) q.push(t->left);
					if (t->right) q.push(t->right);
				}
			}
			return res;
		}
};

dfs方法三

class Solution {
		bool vist[110]= {0};
		vector<int> ans;
	public:
		void dfs(TreeNode *root,int depth) {
			if(!root) return;
			if(!vist[depth]) {
				ans.push_back(root->val);
				vist[depth]=1;
			}
			dfs(root->right,depth+1);
			dfs(root->left,depth+1);
		}

		vector<int> rightSideView(TreeNode* root) {
			dfs(root,1);
			return ans;
		}
};