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].
层次遍历法，找出每层最右真个结点。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// Runtime:7ms
class Solution {
public:
vector<int> rightSideView(TreeNode *root) {
vector<int> nums;
queue<TreeNode*> nodes;
if (root != NULL)
{
nodes.push(root);
}
TreeNode *cur;
while (!nodes.empty())
{
int size = nodes.size();
for (int i = 0; i < size; i++)
{
cur = nodes.front();
nodes.pop();
if (cur->left != NULL)
{
nodes.push(cur->left);
}
if (cur->right != NULL)
{
nodes.push(cur->right);
}
}
nums.push_back(cur->val);
}
return nums;
}
};# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# Runtime:74ms
class Solution:
# @param root, a tree node
# @return a list of integers
def rightSideView(self, root):
nums = []
if root == None:
return nums
nodes = [root]
while nodes:
size = len(nodes)
for i in range(size):
cur = nodes.pop(0)
if cur.left != None:
nodes.append(cur.left)
if cur.right != None:
nodes.append(cur.right)
nums.append(cur.val)
return nums;