LeetCode#102. Binary Tree Level Order Traversal My Submissions Question

102. Binary Tree Level Order Traversal My Submissions Question

Total Accepted: 96217 Total Submissions: 297797 Difficulty: Easy

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}".

解题思路：两个队列

/**
* 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>> Ans;//存放最后结果
if(root==NULL)return Ans;//根节点为空
queue<TreeNode*> Q;
Q.push(root);//根节点进队列
while(!Q.empty()){
vector<int> temp;//存放当前层的值
queue<TreeNode*> tempQ;//存放下一层的结点队列（缓存）
while(!Q.empty()){
TreeNode *head=Q.front();//取队首元素
temp.push_back(head->val);
Q.pop();
if(head->left!=NULL)tempQ.push(head->left);//左孩子非空，进入下一层的结点队列（缓存）
if(head->right!=NULL)tempQ.push(head->right);//右孩子非空，进入下一层的结点队列（缓存）
}
Q=tempQ;//遍历下一层
Ans.push_back(temp);//保存一层结点的值
}
return Ans;
}
};