算法：二叉树倒序逐层返回列表 107. Binary Tree Level Order Traversal II

Posted 2021-09-04 架构师易筋

107. Binary Tree Level Order Traversal II

Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

递归解法

数的遍历，优先用递归，记住每一层的深度即可获取到子数组。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        helper(root, 0, list);
        
        return list;
    }
    
    private void helper(TreeNode root, int depth, List<List<Integer>> list) {
        if (root == null) return;
        if (depth == list.size()) list.add(0, new ArrayList<>());
        list.get(list.size() - depth - 1).add(root.val);
        helper(root.left, depth + 1, list);
        helper(root.right, depth + 1, list);
    }
}