107. Binary Tree Level Order Traversal II(二叉树广度遍历,自底向上输出)
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2022-05-18 19:39:15
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problems:
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]
]
tip:
广度遍历使用队列实现,vector的反转实现自底向上输出。
solution:
迭代法:
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res;
queue<TreeNode *> q{{root}};
while(!q.empty())
{
vector<int> vec;
for(int i=q.size();i>0;i--)
{
TreeNode *t=q.front();
q.pop();
vec.push_back(t->val);
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
res.push_back(vec);
}
reverse(res.begin(),res.end());//reverse()实现vector反转
return res;
}
};
**递归法:**引入变量level来标记当前深度,即层数,
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res;
dfs(root,0,res);
reverse(res.begin(),res.end());
return res;
}
void dfs(TreeNode *root,int level,vector<vector<int>>& res)//这里注意是vector的引用
{
if(!root) return;
if(res.size() == level) res.push_back({});//当level等于向量大小时申请新的一层
res[level].push_back(root->val);
if(root->left) dfs(root->left,level+1,res);
if(root->left) dfs(root->right,level+1,res);
}
};