二叉树层序遍历应用之Binary Tree Right Side View

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]

.

Solutions：

易想到求的是每层中最右的一个节点。

二叉树层序遍历过程中，每次循环开始时，队列中存放的刚好是将要访问的一层的节点。

/**
* 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) {
vector<int> result;
if(root==NULL){
return result;
}
queue<TreeNode*> Q;
Q.push(root);
TreeNode *right;
while(!Q.empty()) {
int Qsize=Q.size();
for(int i=0; i<Qsize; ++i) {
right=Q.front();
Q.pop();
if(right->left != NULL){
Q.push(right->left);
}
if(right->right != NULL){
Q.push(right->right);
}
}
result.push_back(right->val);
}
}
};