06
04/2015
[LeetCode] Binary Tree Right Side View
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：
这道题较为简单，可以用类似于先序遍历的办法来做，但是有点不同，遍历的顺序是根，右，左。用一个map记录每一层第一个出现的值。下面是代码：
/**
* Definition for binary tree
* 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) {
map<int, int> levelToVal; //记录每一层第一个出现的数值。
tranver(root, 0, levelToVal);
vector<int> result;
for(map<int, int>::iterator it=levelToVal.begin(); it!=levelToVal.end(); it++){
result.push_back(it->second);
}
return result;
}
//类似于先序遍历，但是，顺序是上，右，左
void tranver(TreeNode* node, int level, map<int, int>& levelToVal){
if(node==NULL){
return;
}
map<int, int>::iterator it=levelToVal.find(level);
if(it==levelToVal.end()){
levelToVal.insert(map<int, int>::value_type(level, node->val));
}
tranver(node->right, level+1, levelToVal);
tranver(node->left, level+1, levelToVal);
}
};二次刷题2015-10-10
其实可以直接按照线序遍历来做。因为后面的值会覆盖前面的值。
/**
* 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;
map<int, int> m;
helper(root, m, 0);
for(map<int, int>::iterator it = m.begin(); it!=m.end(); it++){
result.push_back(it->second);
}
return result;
}
void helper(TreeNode* root, map<int, int>& m, int level){
if(root == NULL){
return;
}
m[level] = root->val;
helper(root->left, m, level + 1);
helper(root->right, m, level + 1);
}
};
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