Leetcode 102. Binary Tree Level Order Traversal

文章作者：Tyan
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1. Description

2. Solution

• Version 1

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(!root) {
            return result;
        }
        queue<TreeNode*> q;
        q.push(root);
        traverseLevel(q, result);
        return result;
    }
    
private:
    void traverseLevel(queue<TreeNode*>& q, vector<vector<int>>& result) {
        vector<int> values;
        int size = q.size();
        while(size) {
            TreeNode* current = q.front();
            q.pop();
            if(current) {
                values.push_back(current->val);
                if(current->left) {
                    q.push(current->left);
                }
                if(current->right) {
                    q.push(current->right);
                }
            }
            size--;
        }
        result.push_back(values);
        if(!q.empty()) {
            traverseLevel(q, result);
        }
    }
};

• Version 2

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(!root) {
            return result;
        }
        queue<TreeNode*> q1;
        queue<TreeNode*> q2;
        q1.push(root);
        vector<int> values;
        while(!q1.empty()) {
            TreeNode* current = q1.front();
            q1.pop();
            if(current) {
                values.push_back(current->val);
                if(current->left) {
                    q2.push(current->left);
                }
                if(current->right) {
                    q2.push(current->right);
                }
            }
            if(q1.empty()) {
                result.push_back(values);
                values.clear();
                q1 = q2;
                q2 = {};
            }
        }
        return result;
    }
};

Reference

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