Link: https://leetcode.cn/problems/find-leaves-of-binary-tree/
Question
difficulty: mid
adj diff: 4
Given the root of a binary tree, collect a tree's nodes as if you were doing this:
Collect all the leaf nodes.
Remove all the leaf nodes.
Repeat until the tree is empty.
Example 1:
Input: root = [1,2,3,4,5]
Output: [[4,5,3],[2],[1]]
Explanation:
[[3,5,4],[2],[1]] and [[3,4,5],[2],[1]] are also considered correct answers since per each level it does not matter the order on which elements are returned.
Example 2:
Input: root = [1]
Output: [[1]]
Constraints:
The number of nodes in the tree is in the range [1, 100].
-100 <= Node.val <= 100
这个题不简单。主要是:从上往下扫的时候,先看到 root,后看到 leaf,这怎么 build 结果链条啊?
这个思路是:猜测这个 node 在 result list 的哪一层。我们暂且把这个 index 称为 leafLevel。
看代码吧,很短。
Code
class Solution {
public List<List<Integer>> findLeaves(TreeNode root) {
List<List<Integer>> ans = new ArrayList<List<Integer>>();
testDepth(ans, root);
return ans;
}
private int testDepth(List<List<Integer>> ans, TreeNode node) {
if (node == null) return 0;
int leafLevel = 1 + Math.max(testDepth(ans, node.left), testDepth(ans, node.right));
if (ans.size() < leafLevel) {
// because the traverse always bottom-up, so it's safe to add to the end
ans.add(new LinkedList<Integer>());
}
ans.get(leafLevel - 1).add(node.val);
return leafLevel;
}
}