366. Find Leaves of Binary Tree
DFS. Time = O(n), space = O(n)
/**
* 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>> findLeaves(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
getLevel(root, res);
return res;
}
private int getLevel(TreeNode root, List<List<Integer>> res) {
if (root == null) {
return -1;
}
int left = getLevel(root.left, res);
int right = getLevel(root.right, res);
int level = Math.max(left, right) + 1;
if (res.size() <= level) {
res.add(new ArrayList<>());
}
res.get(level).add(root.val);
return level;
}
}