102. 二叉树的层次遍历
给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其层次遍历结果:
[
[3],
[9,20],
[15,7]
]
队列实现
/**
* Definition for a binary tree node.
* 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>> vec; //保存遍历结点
if(root == NULL) return vec;
q.push(root);
while(!q.empty())
{
vector<int> rec; //记录该层次上的结点
for(int i=q.size();i>0;i--)
{
TreeNode* list = q.front();
rec.push_back(list->val);
q.pop();
if(list->left)
{
q.push(list->left);
}
if(list->right)
{
q.push(list->right);
}
}
if(!rec.empty())
{
vec.push_back(rec); //保存当前层的结点
}
}
return vec;
}
};
递归实现
递归获取每一层结点深度,并存入相应数组
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void getTreeNode(TreeNode* node, vector<vector<int>> &vec, int depth) {
if(node != NULL) {
if(vec.size() <= depth){ // 一维长度表示深度
vec.push_back(vector<int>()); // 增加一维长度
}
vec[depth].push_back(node->val);
}
else
return;
getTreeNode(node->left,vec,depth+1);
getTreeNode(node->right,vec,depth+1);
}
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> vec;
getTreeNode(root,vec,0);
return vec;
}
};