Description

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

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]
]

Solution 1(C++)

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;

        if(root == nullptr) return res;

        queue<TreeNode*> layer;
        layer.push(root);

        while(!layer.empty()){
            int layernum = layer.size();
            vector<int> cur;
            while(layernum > 0){
                TreeNode* temp = layer.front();
                cur.push_back(temp->val);
                if(temp->left != nullptr) layer.push(temp->left);
                if(temp->right != nullptr) layer.push(temp->right);
                layer.pop();
                --layernum;
            }
            res.push_back(cur);
        }
        return res;
    }
};

算法分析

这道题考察的是二叉树的层次遍历，之前也有做过树的Easy难度的题目，关键就是层次遍历要利用队列这种数据结构。先进先出。同一层的节点，进入队列，并将当前节点的左右子节点推入队列。即可。

程序分析

注意队列在树的层次遍历中的作用。