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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].
Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.
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分析：
典型的广度优先遍历，每一层的最右边那个元素显然就是当层的右视可见节点！
/**
* 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) {
if(root==NULL)
return result;
queue<TreeNode *> que;
que.push(root);
//广度优先，总是压入每一层最右边的即可
while(!que.empty())
{
int levelNum = que.size();//通过size来判断当层的结束
for(int i=0; i<levelNum; i++)
{
if(que.front()->left != NULL) //先获取该节点下一层的左右子，再获取该节点的元素，因为一旦压入必弹出，所以先处理左右子
que.push(que.front()->left);
if(que.front()->right != NULL)
que.push(que.front()->right);
if(i== (levelNum-1))
result.push_back(que.front()->val);
que.pop();
}
}
return result;
}
private:
vector<int> result;
};
注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！
原文地址：http://blog.csdn.net/ebowtang/article/details/51236061
原作者博客：http://blog.csdn.net/ebowtang
本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895
<LeetCode OJ> 199. Binary Tree Right Side View
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原文地址：http://blog.csdn.net/ebowtang/article/details/51236061
