Leetcode: Binary Tree Level Order Traversal I,II

总结:

1. 注意 cnt-- 的位置以及 level push 的位置

2. II 在 1 的基础上加个 stack 或者直接 reverse 都可以

代码:

class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > res;
		if(root == NULL)
			return res;
		
		queue<TreeNode*> record;
		vector<int> level;
		int cnt = 1, nextcnt = 0;

		record.push(root);
		
		while(!record.empty()) {
			TreeNode * tn = record.front();
			record.pop();
			level.push_back(tn->val);

			if(tn->left) {
				record.push(tn->left);
				nextcnt += 1;
			}
			if(tn->right) {
				record.push(tn->right);
				nextcnt += 1;
			}

			cnt --;
			if(cnt == 0) {
				cnt = nextcnt;
				nextcnt = 0;
				// push_back
				res.push_back(level);
				level.clear();
			}
		
		}
		return res;
    }
};

  

原文地址:https://www.cnblogs.com/xinsheng/p/3467861.html