Leetcode(107) Binary Tree Level Order Traversal II

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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],

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5
  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

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

解法

遍历方法依然是从上往下遍历，但是存的时候需要对列表做一次翻转。

具体代码如下：

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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<TreeNode>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> tmplist = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode tmp = q.poll();
                if (tmp.left != null) {
                    q.offer(tmp.left);
                }
                if (tmp.right != null) {
                    q.offer(tmp.right);
                }
                tmplist.add(tmp.val);
            }
            res.add(tmplist);
        }
        List<List<Integer>> finalres = new ArrayList<>();
        for (int i = res.size() - 1; i >= 0; i--) {
            finalres.add(res.get(i));
        }
        return finalres;
    }
}