leetCode练习(102)
2016-10-28 21:27
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题目:Binary Tree Level Order Traversal
难度:easy
问题描述:
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
return its level order traversal as:
解题思路:按层输出二叉树的所有元素。广度搜索(BFS)的基本题型。基本方法是构造一个储存节点的队列queue,存入根节点root(3),进行BFS时,提出这一层的节点3,再载入3的子节点9和20。此层搜索结束。进入下一层BFS,提出9,输入null,提出20,输入15,7。第二层结束。。。以此循环,直到队列为空结束。
代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res=new ArrayList<>();
Queue<TreeNode> queue=new LinkedList<>();
if(root==null){
return res;
}
queue.add(root);
bfs(res,queue);
return res;
}
private void bfs(List<List<Integer>> res,Queue<TreeNode> queue){
int len=queue.size();
if(len==0){
return;
}
TreeNode t;
List<Integer> list=new ArrayList<>();
for(int i=0;i<len;i++){
t=queue.poll();
list.add(t.val);
if(t.left!=null){
queue.add(t.left);
}
if(t.right!=null){
queue.add(t.right);
}
}
res.add(list);
bfs(res,queue);
}
}
难度:easy
问题描述:
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] ]
解题思路:按层输出二叉树的所有元素。广度搜索(BFS)的基本题型。基本方法是构造一个储存节点的队列queue,存入根节点root(3),进行BFS时,提出这一层的节点3,再载入3的子节点9和20。此层搜索结束。进入下一层BFS,提出9,输入null,提出20,输入15,7。第二层结束。。。以此循环,直到队列为空结束。
代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res=new ArrayList<>();
Queue<TreeNode> queue=new LinkedList<>();
if(root==null){
return res;
}
queue.add(root);
bfs(res,queue);
return res;
}
private void bfs(List<List<Integer>> res,Queue<TreeNode> queue){
int len=queue.size();
if(len==0){
return;
}
TreeNode t;
List<Integer> list=new ArrayList<>();
for(int i=0;i<len;i++){
t=queue.poll();
list.add(t.val);
if(t.left!=null){
queue.add(t.left);
}
if(t.right!=null){
queue.add(t.right);
}
}
res.add(list);
bfs(res,queue);
}
}
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