Binary Tree Level Order Traversal II
Question
- leetcode: Binary Tree Level Order Traversal II | LeetCode OJ
- lintcode: (70) Binary Tree Level Order Traversal II
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).
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]
]
題解
這題在普通的 BFS 基礎上增加了逆序輸出,簡單的實現可以使用 Stack 或者最後對結果逆序。
Java - Stack
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: buttom-up level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (root == null) return result;
Stack<ArrayList<Integer>> s = new Stack<ArrayList<Integer>>();
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
int qLen = q.size();
ArrayList<Integer> aList = new ArrayList<Integer>();
for (int i = 0; i < qLen; i++) {
TreeNode node = q.poll();
aList.add(node.val);
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
s.push(aList);
}
while (!s.empty()) {
result.add(s.pop());
}
return result;
}
}
Java - Reverse
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: buttom-up level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (root == null) return result;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
int qLen = q.size();
ArrayList<Integer> aList = new ArrayList<Integer>();
for (int i = 0; i < qLen; i++) {
TreeNode node = q.poll();
aList.add(node.val);
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
result.add(aList);
}
Collections.reverse(result);
return result;
}
}
源碼分析
Java 中 Queue 是接口,通常可用 LinkedList 實例化。
複雜度分析
時間複雜度爲 , 使用了 Queue 或者 Stack 作爲輔助空間,空間複雜度爲 .