题目:

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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

链接: http://leetcode.com/problems/binary-tree-level-order-traversal-ii/

题解:

这几道题都差不多,有关level order的。也可能有很好的方法,但是都没仔细想。基本思路和其他一样,就是BFS

Time Complexity - O(n), Space Complexity - O(1)。

/**
* 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<>();
if(root == null)
return res;
ArrayList<Integer> list = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0; while(!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
list.add(node.val);
if(node.left != null) {
q.offer(node.left);
nextLevel++;
}
if(node.right != null) {
q.offer(node.right);
nextLevel++;
}
if(curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(0, new ArrayList<Integer>(list));
list.clear();
}
} return res;
}
}

二刷 :

方法跟一刷一样,就是跟普通level order traversal一样,但是保存结果的时候点到以下顺序。

Java:

Time Complexity - O(n), Space Complexity - O(1)。

/**
* 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<>();
if (root == null) {
return res;
}
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0;
List<Integer> level = new ArrayList<>();
while (!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
level.add(node.val);
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(0, new ArrayList<Integer>(level));
level.clear();
}
}
return res;
}
}

题外话: 2/14/2016

时间过得真的很快,不抓紧的话根本刷不完。前面一段时间效率极低,每天一两题,也不愿意花时间刷题,反而停留在舒适区,看看书,看看设计。后来返回来刷题发现根本不熟。

最后祝大家情人节快乐。

三刷:

再看了一下discuss,好像用dfs始终比bfs快.

Java:

/**
* 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<>();
if (root == null) {
return res;
}
List<Integer> level = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0; while (!q.isEmpty()) {
TreeNode node = q.poll();
level.add(node.val);
curLevel--;
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(0, new ArrayList<Integer>(level));
level.clear();
}
}
return res;
}
}

Reference:

https://leetcode.com/discuss/5353/there-better-regular-level-order-traversal-reverse-result

https://leetcode.com/discuss/91100/share-java-2ms-recursive-solution

https://leetcode.com/discuss/22538/my-dfs-and-bfs-java-solution

https://leetcode.com/discuss/81189/java-1ms-beats-98%25-using-preorder

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