leetcode 102 Binary Tree Level Order Traversal（难易度：Easy）

Binary Tree Level Order Traversal

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},

    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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

代码：

//解法一：（递归）
<pre name="code" class="cpp">/**

* 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) {traverse(root, 1, result);return result;}void traverse(struct TreeNode *root, size_t level, vector<vector<int>> &result) {if (root == NULL)return;if (level > result.size())result.push_back(vector<int>());result[level - 1].push_back(root->val);traverse(root->left, level + 1, result);traverse(root->right, level + 1, result);}private:vector<vector<int>> result;};

//解法二：（迭代）
/**
 * 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) {
        if (root == nullptr)
	    return result;
	queue<TreeNode *> current,  //当前层的节点
			  next;	    //下一层的节点
	vector<int> level;

	current.push(root);
	while (!current.empty()) {
	    while (!current.empty()) {
		 TreeNode *node = current.front();
		 current.pop();
		 level.push_back(node->val);
		 if (node->left != nullptr) next.push(node->left);
		 if (node->right != nullptr)	next.push(node->right);
	    }	//当前层节点全部处理完
	    result.push_back(level);
            level.clear();
	    swap(next, current);
	}
	return result;
    }
private:
    vector<vector<int>> result;
};

原题地址：https://leetcode.com/problems/binary-tree-level-order-traversal/