Binary Tree Level Order Traversal II
描述
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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7]
[9,20],
[3],
]
分析
在上一题 Binary Tree Level Order Traversal 的基础上，reverse()一下即可。
递归版
// Binary Tree Level Order Traversal II
// 递归版，时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
traverse(root, 1, result);
Collections.reverse(result);
return result;
}
void traverse(TreeNode root, int level,
List<List<Integer>> result) {
if (root == null) return;
if (level > result.size())
result.add(new ArrayList<>());
result.get(level-1).add(root.val);
traverse(root.left, level+1, result);
traverse(root.right, level+1, result);
}
}
迭代版
// Binary Tree Level Order Traversal II
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
Queue<TreeNode> current = new LinkedList<>();
Queue<TreeNode> next = new LinkedList<>();
if(root == null) {
return result;
} else {
current.offer(root);
}
while (!current.isEmpty()) {
ArrayList<Integer> level = new ArrayList<>(); // elments in one level
while (!current.isEmpty()) {
TreeNode node = current.poll();
level.add(node.val);
if (node.left != null) next.add(node.left);
if (node.right != null) next.add(node.right);
}
result.add(level);
// swap
Queue<TreeNode> tmp = current;
current = next;
next = tmp;
}
Collections.reverse(result);
return result;
}
}