107 Binary Tree Level Order Traversal II
107. Binary Tree Level Order Traversal II
1. Question
Given a binary tree, return thebottom-up level ordertraversal 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]
]
2. Implementation
(1) BFS
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
List<Integer> level = null;
while (!queue.isEmpty()) {
int size = queue.size();
level = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode curNode = queue.remove();
level.add(curNode.val);
if (curNode.left != null) {
queue.add(curNode.left);
}
if (curNode.right != null) {
queue.add(curNode.right);
}
}
res.add(0, level);
}
return res;
}
}
(2) DFS
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
getLevelOrderByDFS(root, 0, res);
return res;
}
public void getLevelOrderByDFS(TreeNode node, int depth, List<List<Integer>> res) {
if (node == null) {
return;
}
if (res.size() == depth) {
res.add(0, new ArrayList<>());
}
res.get(res.size() - depth - 1).add(node.val);
getLevelOrderByDFS(node.left, depth + 1, res);
getLevelOrderByDFS(node.right, depth + 1, res);
}
}
3. Time & Space Complexity
BFS: 时间复杂度: O(n), 空间复杂度: O(n)
DFS: 时间复杂度: O(n), 空间复杂度: O(n)
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