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
要将原来的TreeNode增加一个next指针 指向其同层的下一个 很明显地想到是用层序遍历的方法 代码如下:
/**
* 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 ;
int count=1;
int level=0;
Queue<TreeLinkNode> que =new LinkedList<TreeLinkNode>();
que.offer(root);
while(que.isEmpty()!=true){
level=0;
for(int i=0;i<count;i++){
root=que.peek();
que.poll();
if(i<count-1){
root.next=que.peek();
}else{
root.next=null;
}
if(root.left!=null){
que.offer(root.left);
que.offer(root.right);
level++;
level++;
}
}
count=level;
}
}
}
原来题中要求的是空间复杂度为0(1) 但遍历的话空间复杂度为O(log N)上面方法不满足要求 从例子来看 每一层的next设置都可以根据自身和上一层的结果来设置 所以采用递归即可 代码如下:
public class Solution {
public void connect(TreeLinkNode root) {
if(root==null)return ;
if(root.left!=null){
root.left.next=root.right;
}
if(root.right!=null&&root.next!=null){
root.right.next=root.next.left;
}else{
if(root.right!=null){
root.right.next=null;
}
}
connect(root.left);
connect(root.right);
}
}