107. Binary Tree Level Order Traversal II (E)

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Binary Tree Level Order Traversal II (E)

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],

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

题意

按从下到上的顺序记录二叉树的每一层。

思路

BFS后再将结果集逆序。

代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        Queue<TreeNode> q = new ArrayDeque<>();

        if (root != null) {
            q.offer(root);
        }

        while (!q.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = q.poll();
                list.add(cur.val);
                if (cur.left!=null) q.offer(cur.left);
                if (cur.right!=null) q.offer(cur.right);
            }
            ans.add(list);
        }
        
        Collections.reverse(ans);
        return ans;
    }
}