leetcode解题: Binary Tree Level Order Traversal (102)

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Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

解法1：One Queue

比较基础的BST算法，主要是掌握用一个queue和一个每层的计数器k来维护当前层的node个数
C++

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

Java

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