Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
改变树的结构。
第一种用队列,并不是很快。
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) { if( root == null)
return ;
Queue queue = new LinkedList<TreeLinkNode>(); queue.add(root); while( !queue.isEmpty() ){ TreeLinkNode node = (TreeLinkNode) queue.poll(); if( node.left != null){
node.left.next = node.right;
if( node.next != null )
node.right.next = node.next.left;
queue.add(node.left);
queue.add(node.right);
}
} }
}
不使用队列就会达到最快。
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if( root == null)
return ;
TreeLinkNode node1 = root,node2 = root;
while( node2.left != null ){ node1 = node2.left;
while( node2.next != null ){
node2.left.next = node2.right;
node2.right.next = node2.next.left;
node2 = node2.next;
}
node2.left.next = node2.right;
node2 = node1; } }
}