Given a binary tree, return thezigzag level ordertraversal 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]
]
2. Implementation
(1) BFS
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
boolean reversed = false;
List<Integer> level = null;
int depth = 0;
while (!queue.isEmpty()) {
int size = queue.size();
level = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode curNode = queue.remove();
if (depth % 2 == 0) {
level.add(curNode.val);
}
else {
level.add(0, curNode.val);
}
if (curNode.left != null) {
queue.add(curNode.left);
}
if (curNode.right != null) {
queue.add(curNode.right);
}
}
res.add(level);
++depth;
}
return res;
}
}
(2) DFS
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
getZigzagLevelOrderByDFS(root, 0, res);
return res;
}
public void getZigzagLevelOrderByDFS(TreeNode node, int depth, List<List<Integer>> res) {
if (node == null) {
return;
}
if (res.size() == depth) {
res.add(new ArrayList<>());
}
if (depth % 2 == 0) {
res.get(depth).add(node.val);
}
else {
res.get(depth).add(0, node.val);
}
getZigzagLevelOrderByDFS(node.left, depth + 1, res);
getZigzagLevelOrderByDFS(node.right, depth + 1, res);
}
}