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},

return its level order traversal as:

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

题目翻译:
给定一颗二叉树，返回一个二维数组，使这个二维数组的每一个元素代表着二叉树的一层的元素.例子已经明确给出.

题目分析:
对于二叉树的问题，我们第一想到的就是DFS或者BFS, DFS更易于理解代码，如果处理数据量不是很大的话.对于这样的面试题，我建议用DFS来求解.

需要注意的点为:

对于一棵树，如果我们要求每一层的节点，并且存在一个二维数组里，首先我们要建一个二维数组，但是这个二维数组建多大的合适呢？我们就要求出这颗树的深度，根据深度来建立二维数组.
题目要求为从左往右添加,所以我们也就是要先放左边的节点，再放右边的节点.
对于这道题，我们首先就是要用DFS来求出这颗树的高度，之后再用DFS对于每一层遍历，这样节省了空间复杂度.

时间复杂度分析:
由于两次DFS是并列的，并没有嵌套，所以我们的时间复杂度为O(n).

代码如下:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
/* for this question, we need to construct the ret vector first
   thus, we need to know the depth of this tree, we write a simple
   function to calculate the height of this tree */
    vector<vector<int> > levelOrder(TreeNode *root) {
       int depth = getHeight(root);
       vector<vector<int>> ret(depth);
       if(depth == 0) //invalid check
            return ret;
        getSolution(ret,root,0);
        return ret;
    }
    void getSolution(vector<vector<int>>& ret, TreeNode* root, int level)
    {
        if(root == NULL)
            return;
        ret[level].push_back(root->val);
        getSolution(ret,root->left,level+1);
        getSolution(ret,root->right,level+1);
    }
    int getHeight(TreeNode* root)
    {
        if(root == NULL)
            return 0;
        int left = getHeight(root->left);
        int right = getHeight(root->right);
        int height = (left > right?left:right)+1;
        return height;
    }
};

Binary Tree Level Order Traversal II

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

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

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

题目翻译:
给定一颗二叉树，返回一个二维数组，这个二维数组要满足这个条件，二维数组的第一个一维数组要是这可二叉树的最下面一层，之后以此类推，根据以上例子应该知道要求的条件。

题目分析 && 解题思路:
这道题和Binary Tree Level Order Traversal 几乎是一摸一样的，唯一不同的就是二维数组的存储顺序，详见以下代码.

时间复杂度:
O(n)-树的dfs均为O(n)

代码如下:


/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        int depth = height(root);
        vector<vector<int>> ret(depth);
        if(depth == 0)
            return ret;
        DFS(ret,ret.size()-1, root);
        return ret;
    }
    void DFS(vector<vector<int>>& ret, int level, TreeNode* root)
    {
        if(root == NULL)
            return;
        ret[level].push_back(root->val);
        DFS(ret,level-1,root->left);
        DFS(ret,level-1,root->right);
    }
    /* get the height first of all */
    int height(TreeNode* root)
    {
        if(root == NULL)
            return 0;
        int left_side = height(root->left);
        int right_side = height(root->right);
        int height = (left_side > right_side?left_side:right_side)+1;
        return height;
    }
};

Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]

如果完成了上面两题，这题应该是很简单的，我们只需要将得到的数据按照zigzag的方式翻转一下，代码如下：

class Solution {
public:
vector<vector<int> > vals;
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        build(root, 1);
        //翻转
        for(int i = 1; i < vals.size(); i+=2) {
            reverse(vals[i].begin(), vals[i].end());
        }
        return vals;
    }
    void build(TreeNode* node, int level) {
        if(!node) {
            return;
        }
        if(vals.size() <= level - 1) {
            vals.push_back(vector<int>());
        }
        vals[level - 1].push_back(node->val);
        if(node->left) {
            build(node->left, level + 1);
        }
        if(node->right) {
            build(node->right, level + 1);
        }
    }
};