March 19, 2022

366. Find Leaves of Binary Tree

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;
    }
}
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