[LC] 107. Binary Tree Level Order Traversal II
2024-09-02 23:39:08
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
] Solution 1:
BFS
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> arrList = new ArrayList<>();
if (root == null) {
return arrList;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
list.add(cur.val);
if (cur.left != null) {
queue.offer(cur.left);
}
if (cur.right != null) {
queue.offer(cur.right);
}
}
arrList.add(0, list);
}
return arrList;
}
}
Solution 2:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> arrList = new ArrayList<>();
helper(arrList, 0, root);
return arrList;
} private void helper(List<List<Integer>> res, int depth, TreeNode root) {
if (root == null) {
return;
}
// backtrack case, resSize >= depth
if (depth >= res.size()) {
res.add(0, new LinkedList<>());
}
res.get(res.size() - 1 - depth).add(root.val);
helper(res, depth + 1, root.left);
helper(res, depth + 1, root.right);
}
}
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