102. 二叉树的层次遍历 - Binary Tree Level Order Traversal
【层次遍历-迭代】 【层次遍历-递归】
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)。
Example:
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]
]
Solutions
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
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Solution 【层次遍历-迭代】 ( 2ms)
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class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<>();
List<List<Integer>> result = new LinkedList<>();
if(root == null)
return result;
queue.offer(root);
TreeNode current, left, right;
int size = 0;
List<Integer> curList;
while(queue.isEmpty() == false){
size = queue.size();
curList = new LinkedList<>();
while(size != 0){
current = queue.poll();
size--;
curList.add(current.val);
left = current.left;
right = current.right;
if(left != null)
queue.add(left);
if(right != null)
queue.add(right);
}
result.add(curList);
}
return result;
}
}
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Solution 【层次遍历-递归】 ( 0ms)
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class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
function(root,list,0);
return list;
}
public void function(TreeNode node,List<List<Integer>> list,int level){
if(node == null)
return;
if(list.size()<=level)
list.add(new ArrayList<Integer>());
list.get(level).add(node.val);
function(node.left,list,level+1);
function(node.right,list,level+1);
}
}
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Notes