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].
Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.
层次遍历，到每一层最后一个节点，即装入ret
每层最后一个节点的判断：
（1）队列为空
（2）下一个待遍历节点在下一层
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct Node
{
TreeNode* tnode;
int level; Node(TreeNode* t, int l): tnode(t), level(l) {}
}; class Solution {
public:
vector<int> rightSideView(TreeNode *root) {
vector<int> ret;
if(root == NULL)
return ret;
queue<Node*> q;
int curLevel = ;
Node* rootNode = new Node(root,);
q.push(rootNode); while(!q.empty())
{
Node* front = q.front();
q.pop(); if(q.empty() || q.front()->level > front->level)
//last node of current level
ret.push_back(front->tnode->val); if(front->tnode->left)
{
Node* leftNode = new Node(front->tnode->left, front->level+);
q.push(leftNode);
} if(front->tnode->right)
{
Node* rightNode = new Node(front->tnode->right, front->level+);
q.push(rightNode);
}
}
return ret;
}
};