[LeetCode]Binary Tree Level Order Traversal

题目描述:(链接)

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,#,#,15,7},

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解题思路：

广度优先遍历

递归版：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         travel(root, 1);
14         return result;
15     }
16     
17     void travel(TreeNode *root, int level) {
18         if (!root) return;
19         if (level > result.size()) {
20             result.push_back(vector<int>());
21         }
22         
23         result[level - 1].push_back(root->val);
24         travel(root->left, level + 1);
25         travel(root->right, level + 1);
26     }
27 private:
28     vector<vector<int>> result;
29 };

迭代版:

借助一个队列实现先进先出：

/**
 * 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 *> current, next;
        vector<int> level;
        current.push(root);
        
        while (!current.empty()) {
            while (!current.empty()) {
                TreeNode *tmp = current.front();
                current.pop();
                level.push_back(tmp->val);
                
                if (tmp->left != nullptr) next.push(tmp->left);
                if (tmp->right != nullptr) next.push(tmp->right);
            }
            
            result.push_back(level);
            level.clear();
            swap(current, next);
        }
        return result;
    }
};