Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
- You may only use constant extra space.
For example,
Given the following binary tree,
1 / 2 3 / \ 4 5 7
After calling your function, the tree should look like:
1 -> NULL / 2 -> 3 -> NULL / \ 4-> 5 -> 7 -> NULL
解题思路一:
本题的难点在于需要逐层遍历才行,因此可以用Java for LeetCode 102 Binary Tree Level Order Traversal的思路,JAVA实现如下:
public void connect(TreeLinkNode root) { if (root == null || (root.left == null && root.right == null)) return; List<List<TreeLinkNode>> list=new ArrayList<List<TreeLinkNode>>(); Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>(); queue.add(root); while (queue.size() != 0) { List<TreeLinkNode> alist = new ArrayList<TreeLinkNode>(); for (TreeLinkNode child : queue) alist.add(child); list.add(new ArrayList<TreeLinkNode>(alist)); Queue<TreeLinkNode> queue2=queue; queue=new LinkedList<TreeLinkNode>(); for(TreeLinkNode child:queue2){ if (child.left != null) queue.add(child.left); if (child.right != null) queue.add(child.right); } } for(List<TreeLinkNode> alist:list) for(TreeLinkNode aNode:alist) connectARoot(aNode); } public static void connectARoot(TreeLinkNode root){ if (root == null || (root.left == null && root.right == null)) return; if (root.next == null) { if (root.left != null) root.left.next = root.right; } else if (root.right == null) { TreeLinkNode temp=root.next; while(temp!=null){ if(temp.left==null&&temp.right==null) temp=temp.next; else break; } if(temp!=null) root.left.next = (temp.left == null?temp.right:temp.left); }else { if (root.left != null) root.left.next = root.right; TreeLinkNode temp=root.next; while(temp!=null){ if(temp.left==null&&temp.right==null) temp=temp.next; else break; } if(temp!=null) root.right.next = (temp.left == null?temp.right:temp.left); } }
解题思路二:
不使用队列,请移步Populating Next Right Pointers in Each Node I [email protected]
时间: 2024-10-27 01:09:52