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;
    }
};