leetcode@ [199] Binary Tree Right Side View (DFS/BFS)

Posted 2020-06-21 流白

https://leetcode.com/problems/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].

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class node {
    public: 
        TreeNode *nd;
        int lv;
        node(TreeNode *rhs, int l): nd(rhs), lv(l) {}
};

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<pair<int, int> > load;
        vector<int> res;
        if(root == NULL)  return res;
        
        stack<node> q;
        q.push(node(root, 0));
        int lv = 0;
        
        while(!q.empty()) {
            node top = q.top();
            int cur_lv = top.lv;
            q.pop();
            load.push_back(make_pair(top.nd->val, cur_lv));
            
            if(top.nd->left)  q.push(node(top.nd->left, cur_lv+1));
            if(top.nd->right)  q.push(node(top.nd->right, cur_lv+1));
        }
        
        int elv = 0;
        for(int i=0; i<load.size(); ++i) {
            if(elv == load[i].second) {
                res.push_back(load[i].first);
                ++elv;
            }
        }
        
        return res;
    }
};