binary-tree-level-order-traversal

gistfile1.txt

https://oj.leetcode.com/problems/binary-tree-level-order-traversal/

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
 /*
 solution
 1. BFS + 매 iteration마다 marker를 queue에 삽입하여 level  구분하기
 2. counter를 이용하여 매 iteration마다 자식노드 갯수 세기
 3. in-order로 depth level을 알수있으므로 바로 push(loop로 구현할 경우  in-place로 구현 가능)
 */
 
class Solution {
    public:
    vector<vector<int>> levelOrder(TreeNode *root) {
        vector<vector<int>> res;

        if(root == NULL) return res;
        
        vector<int> level;
        queue<TreeNode*> input;
        int cnt = 1;
        input.push(root);
      
        while(!input.empty()) {
            level.clear();
            int levelCnt = 0;
            
            for(int i=0 ; i<cnt; i++) {
                TreeNode* node = input.front();
                level.push_back(node->val);
                input.pop();
                
                if(node->left != NULL) {
                    input.push(node->left);
                    levelCnt++;
                }
                
                if(node->right != NULL) {
                    input.push(node->right);
                    levelCnt++;
                }
            }
            
            cnt = levelCnt;
            res.push_back(level);
        }
    }
};