Binary Tree Level Order Traversal 、Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal

题目描述：

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]
]

解题思路:

求一棵二叉树的各层节点，并且按层存储在二元数组中。利用BFS即可实现。我的博客《二元树的生成、遍历、以及最短最长路径查询》中已经提到，利用队列可实现二元树的横向遍历，这本题中，我们依旧利用队列对树进行BFS，使用两个队列，q表示本层节点，q1表示下层节点，依次进队列，出队列，便可构造二维向量。
代码如下：

class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
vector<int> temp;
if(root==NULL) return res;
queue<TreeNode*> q;
q.push(root);
TreeNode *node;
while(!q.empty())
{
queue<TreeNode*> q1;
while(!q.empty())
{
node=q.front();
temp.push_back(node->val);
q.pop();
if(node->left!=NULL)
{
q1.push(node->left);
}
if(node->right!=NULL)
{
q1.push(node->right);
}

}
res.push_back(temp);
q=q1;
temp.clear();
}
return res;
}
};

Binary Tree Level Order Traversal II

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

For example:

Given binary tree 

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

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

解题思路：

同Binary Tree Level Order Traversal一样，最后将二元向量翻转即可，代码如下：

class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res;
vector<int> temp;
if(root==NULL) return res;
queue<TreeNode*> q;
q.push(root);
TreeNode* node;
while(!q.empty())
{
queue<TreeNode*> q1;
while(!q.empty())
{
node=q.front();
q.pop();
temp.push_back(node->val);
if(node->left!=NULL)
q1.push(node->left);
if(node->right!=NULL)
q1.push(node->right);
}
res.push_back(temp);
temp.clear();
q=q1;
}
reverse(res.begin(),res.end());
return res;
}
};