[LeetCode] 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].

解题思路

层次遍历法,找出每层最右真个结点。

实现代码1

/**
* 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;
}
};

实现代码2

# 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;

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