【leetcode】Binary Tree Level Order Traversal I & II

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]
]

分层遍历二叉树，用BFS即可

class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> Q;
vector<vector<int>> ans;
Q.push(root);
while(!Q.empty())
{
int e = Q.size(); //当前层的结束位置
vector<int> partans;
for(int i = 0; i < e; i++)
{
TreeNode * p = Q.front();
Q.pop();
if(NULL != p)
{
partans.push_back(p->val);
Q.push(p->left);
Q.push(p->right);
}
}
if(!partans.empty())
ans.push_back(partans);
}
return ans;
}
};

网上DFS的代码：

class Solution {
protected:
vector<vector<int>> ans;
void dfs(TreeNode *root, int height){
if (root == NULL)
return;
while (ans.size() <= height)
ans.push_back(vector<int>());
ans[height].push_back(root->val);
dfs(root->left, height + 1);
dfs(root->right, height + 1);
}

public:
vector<vector<int>> levelOrder(TreeNode* root) {
dfs(root, 0);
return ans;
}
};

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]
]

跟上面只是顺序变了，加个reverse就行了。