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]
]
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},
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 7return 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);
}
}
};