102. 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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
107. 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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
102和107的解题思路就是层次遍历,但是一个是从头添加一个是从尾添加,只有一句话不同:
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>>res;
queue<TreeNode*>level;
if(root)
level.push(root);
while(!level.empty())
{
vector<int>vec;
int n=level.size();
for(int i=0;i<n;i++)
{
TreeNode *node=level.front();
//这句话一定要写在for循环中,因为弹出之后,队列末尾的元素变了。
if(node->left)
level.push(node->left);
if(node->right)
level.push(node->right);
vec.push_back(node->val);
level.pop();
}
res.push(vec);//102
res.insert(res.begin(),vec);//107
}
return res;
}
};