Binary Tree Zigzag Level Order Traversal
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,null,null,15,7],
3
/ \ 9 20 / \ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
Solution
/**
* 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>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
List<TreeNode> currentLevel = new ArrayList<>();
if (root != null) {
currentLevel.add(root);
}
boolean left = false;
while (!currentLevel.isEmpty()) {
List<Integer> cur = currentLevel.stream().map(item -> item.val).collect(Collectors.toList());
result.add(cur);
List<TreeNode> nextLevel = new ArrayList<>();
if (left) {
for(int i = currentLevel.size() - 1; i >= 0; i --) {
TreeNode curNode = currentLevel.get(i);
if (curNode.left != null) {
nextLevel.add(curNode.left);
}
if (curNode.right != null) {
nextLevel.add(curNode.right);
}
}
left = false;
} else {
for(int i = currentLevel.size() - 1; i >= 0; i --) {
TreeNode curNode = currentLevel.get(i);
if (curNode.right != null) {
nextLevel.add(curNode.right);
}
if (curNode.left != null) {
nextLevel.add(curNode.left);
}
}
left = true;
}
currentLevel = nextLevel;
}
return result;
}
}