题解 | #求二叉树的层序遍历#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > levelOrder(TreeNode* root) {
        // write code here
        vector<vector<int> > res;
        if(root == nullptr) return res;
        TreeNode* cur = root;
        TreeNode* curEnd = root;		//当前层的最后一个节点
        TreeNode* nextEnd = nullptr;		//标记下一层节点的最右节点
        queue<TreeNode*> q;
        q.push(root);
        vector<int> levelRes;		//标记每一层的节点
        while(q.size()){
            cur = q.front();
            q.pop();
            if(cur->left){
                q.push(cur->left);
                nextEnd = cur->left;
            }
            if(cur->right){
                q.push(cur->right);
                nextEnd = cur->right;
            }
            if(cur == curEnd){		//到了当前层的最右节点，则结算，把结果添加入res，并刷新curEnd和nextEnd
                levelRes.push_back(cur->val);
                res.push_back(levelRes);
                levelRes.clear();
                curEnd = nextEnd;
                nextEnd = nullptr;
            }else{
                levelRes.push_back(cur->val);		//记录当前层节点
            }
        }
        return res;
    }
};