LeetCode 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]
]

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.
题意：给出一个二叉树，求其的层次遍历
思路：用BFS方法，在每一层后加null，从队列中取出一个，如果不为null，就认为是同一层，直接加到该层的最后，如果为null，说明是一层的结束，此时队列为空，就退出，不为空，将null也入队列，再新建一层

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<List<Integer>>();
ArrayList<Integer> level = null;

Queue<TreeNode> q = new LinkedList<TreeNode>();

if (root != null) {
q.add(root);
q.add(null);
}

level = new ArrayList<Integer>();

while (!q.isEmpty()) {
TreeNode cur = q.poll();
if (cur == null) {
ans.add(level);
if (q.isEmpty()) break;
q.add(null);
level = new ArrayList<Integer>();
continue;
}

level.add(cur.val);
if (cur.left != null) {
q.add(cur.left);
}

if (cur.right != null) {
q.add(cur.right);
}
}
return ans;
}
}

另一种思路：用DFS方法，分别求左、右子树的层次遍历结果，再合并

/**
* 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> > levelOrder(TreeNode *root) {
vector<vector<int> > vv;
vector<int> v;

if (root) {
v.push_back(root->val);
vv.push_back(v);
}

vector<vector<int> > vvleft, vvright;

if (root && root->left) {
vvleft = levelOrder(root->left);
}

if (root && root->right) {
vvright = levelOrder(root->right);
}

for (int i = 0; i < vvleft.size() || i < vvright.size(); i++) {
if (i < vvleft.size() && i < vvright.size()) {
vvleft[i].insert(vvleft[i].end(), vvright[i].begin(), vvright[i].end());
vv.push_back(vvleft[i]);
} else if (i < vvleft.size()) {
vv.push_back(vvleft[i]);
} else {
vv.push_back(vvright[i]);
}
}

return vv;
}
};

第三种解：用BFS，先将根结点入队列，在队列操作中，取出一个结点，如果非空，将左、右树入队列（不管是不是空），同时将入队列的结点放入一个数组中。接着是对这个数组操作，在扫描当前层时，可以计算下一层的结束下标，以此类推

/**
* 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> > levelOrder(TreeNode *root) {
queue<TreeNode*> q;
vector<vector<int> > vv;
vector<int> v;

if (root) {
v.push_back(root->val);
vv.push_back(v);
}

q.push(root);
vector<TreeNode*> vt;
while (!q.empty()) {
TreeNode* cur = q.front(); q.pop();
vt.push_back(cur);
if (cur == NULL) continue;
q.push(cur->left);
q.push(cur->right);
}

int step = 2;
size_t j;

for (size_t i = 1; i < vt.size(); i = j) {
v.clear();

int cnt = 0;

for (j = i; j < i + step && j < vt.size(); j++) {
if (vt[j]) {
v.push_back(vt[j]->val);
cnt += 2;
}
}

step = cnt;
if (v.size() > 0) {
vv.push_back(v);
}
}

return vv;
}
};