一.题目描述
给出一棵二叉树,返回其节点值的层次遍历(逐层从左往右访问)
样例
给一棵二叉树 {3,9,20,#,#,15,7}
:
3
/ \
9 20
/ \
15 7
返回他的分层遍历结果:
[
[3],
[9,20],
[15,7]
]
二.解题思路
给一棵二叉树 {3,9,20,#,#,15,7}
:
3
/ \
9 20
/ \
15 7
返回他的分层遍历结果:
[
[3],
[9,20],
[15,7]
]
利用队列的先进先出,先让根节点入队,然后依次出队保存在向量中,同时让其左右儿子入队,层层进行完成层次遍历.
三.实现代码
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
/**
* @param root: The root of binary tree.
* @return: Level order a list of lists of integer
*/
public:
vector<vector<int>> levelOrder(TreeNode *root) {
// write your code here
vector<vector<int>>lever;
if(root==NULL) return lever;
vector<int>dlever;
queue<TreeNode*>q;
q.push(root);
int i=1;
int j=0;
while(!q.empty()){
TreeNode *p=q.front();
q.pop();
if(p==NULL) {
++j;
}
else {
dlever.push_back(p->val);
q.push(p->left);
q.push(p->right);
}
if(i==(dlever.size()+j)&&dlever.size()!=0){
lever.push_back(dlever);
dlever.clear();
i=i*2;
j=j*2;
}
}
return lever;
}
};
四.感悟
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
/**
* @param root: The root of binary tree.
* @return: Level order a list of lists of integer
*/
public:
vector<vector<int>> levelOrder(TreeNode *root) {
// write your code here
vector<vector<int>>lever;
if(root==NULL) return lever;
vector<int>dlever;
queue<TreeNode*>q;
q.push(root);
int i=1;
int j=0;
while(!q.empty()){
TreeNode *p=q.front();
q.pop();
if(p==NULL) {
++j;
}
else {
dlever.push_back(p->val);
q.push(p->left);
q.push(p->right);
}
if(i==(dlever.size()+j)&&dlever.size()!=0){
lever.push_back(dlever);
dlever.clear();
i=i*2;
j=j*2;
}
}
return lever;
}
};
队列的用法掌握得不好这道题就变得有点难.