Easy and understandable solution in Java


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    In "dfs" method, "right" is exactly the neighbor of "cur", end of story!

    public class Solution {
        public void connect(TreeLinkNode root) {
            dfs(root, null);
        }
        void dfs(TreeLinkNode cur, TreeLinkNode right){
            if(cur == null) return;
            cur.next = right;
            dfs(cur.left, cur.right);
            if(right == null){
                dfs(cur.right, null);
            }else{
                dfs(cur.right, right.left);
            }
        }
    }
    

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