[LeetCode] 199. Binary Tree Right Side View

轉自LeetCode

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].

<Solution>
這題可以看做 Binary Tree Level Order Traversal 的衍生題

利用同樣思路解即可，主要變動的地方是

• 每次 iteration，把 queue 裡面的最後一個 node，放到答案裡面

code 如下

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	public:
	vector<int> rightSideView(TreeNode* root) {
	vector<int> ans;
	if(!root) {
	return ans;
	}
	queue<TreeNode *> nodeQue;
	nodeQue.push(root);
	while(!nodeQue.empty()) {
	//>> put the right most node to ans
	ans.push_back(nodeQue.back()->val);
	//>> level order traversal
	int len = nodeQue.size();
	for(int i = 0; i < len; i++) {
	TreeNode *tmpNode = nodeQue.front();
	nodeQue.pop();
	if(tmpNode->left) {
	nodeQue.push(tmpNode->left);
	}
	if(tmpNode->right) {
	nodeQue.push(tmpNode->right);
	}
	}
	}
	return ans;
	}
	};

view raw binaryTreeRightSide.cpp hosted with ❤ by GitHub