BinaryTreeLevelOrderTraversalII [source code]

public class BinaryTreeLevelOrderTraversalII {
static

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Deque<List<Integer>> levels = new ArrayDeque<>();
        List<Integer> level = new ArrayList<>(Arrays.asList(root.val));
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            level = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
            levels.push(level);
        }
        res.addAll(levels);
        return res;
    }
}

    public static void main(String[] args) {
        BinaryTreeLevelOrderTraversalII.Solution tester = new BinaryTreeLevelOrderTraversalII.Solution();
    }
}

虽然其实是一个比较基础的题目, 但是 BFS 还真是第一次写, 有些细节确实是第一次碰到和琢磨;

LinkedList as Queue, not ArrayList

这个也是一个小盲点;

BFS操作的核心就是每一个 iteration 解决一个 level; 可以说有一个invariant就是there are never nodes belonging to more than one levels in the queue at the same time;

inner header 的写法:

for (int i = 0; i < size; i++) {

不能直接用 queue.size(), 因为每一个 iteration, queue 的 size 都是实时变化的. size 这里存的其实是当前 iteration 正在处理的 level 的 size;

最后的速度是3(25), 不算好;

Problem Description

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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
Difficulty:Easy
Category:Algorithms
Acceptance:39.62%
Contributor: LeetCode
Related Topics
tree breadth-first search
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Binary Tree Level Order Traversal