Binary Tree Zigzag Level Order Traversal
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Problem:
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / \ 9 20 / \ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7]]
改变入队规则,把当前被访问节点的孩子节点插入到下一层队列的队首,而不是整个队列的队尾,如果被访问的是奇数层,则先插入左节点,再插入右节点,如果被访问的是偶数层,则先插入右节点,再插入左节点。
Solution:
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if(root==null)
return result;
List<TreeNode> qList = new ArrayList<>();
qList.add(root);
while(!qList.isEmpty())
{
int size = qList.size();
ArrayList<Integer> list = new ArrayList<>();
while(size-->0)
{
TreeNode t = qList.remove(0);
list.add(t.val);
if(result.size()%2==0)
{
if(t.left!=null)
qList.add(size,t.left);
if(t.right!=null)
qList.add(size,t.right);
}
else
{
if(t.right!=null)
qList.add(size,t.right);
if(t.left!=null)
qList.add(size,t.left);
}
}
result.add(list);
}
return result;
}
}
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if(root==null)
return result;
List<TreeNode> qList = new ArrayList<>();
qList.add(root);
while(!qList.isEmpty())
{
int size = qList.size();
ArrayList<Integer> list = new ArrayList<>();
while(size-->0)
{
TreeNode t = qList.remove(0);
list.add(t.val);
if(result.size()%2==0)
{
if(t.left!=null)
qList.add(size,t.left);
if(t.right!=null)
qList.add(size,t.right);
}
else
{
if(t.right!=null)
qList.add(size,t.right);
if(t.left!=null)
qList.add(size,t.left);
}
}
result.add(list);
}
return result;
}
}
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