Level Order Traversal II
出处
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]
]
Solution
Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.
Complexity
时间复杂度 O(n),空间复杂度 O(n)
Code
/**
* 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>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (root == null) return res;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(null);
List<Integer> level = new ArrayList<Integer>();
while(true) {
TreeNode node = q.poll();
if (node != null) {
level.add(node.val);
if(node.left!=null) q.offer(node.left);
if(node.right!=null) q.offer(node.right);
} else {
res.add(level);
level = new ArrayList<Integer>();
if(q.isEmpty()==true) break;
q.offer(null);
}
}
Collections.reverse(res);
return res;
}
}