LeetCode——Populating Next Right Pointers in Each Node II
2014-08-04 21:35
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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,
After calling your function, the tree should look like:
原题链接:https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
题目:前一题的继续,如果给定的是任意的二叉树,前题的代码还能用吗?
思路:显然,前一题不能用了。只好用层序遍历。
public void connect(TreeLinkNode root) {
if(root == null)
return;
TreeLinkNode parent = root,pre,next;
while(parent != null){
pre = null;
next = null;
while(parent != null){
if(next == null)
next = (parent.left == null) ? parent.right : parent.left;
if(parent.left != null){
if(pre != null){
pre.next = parent.left;
pre = pre.next;
}else
pre = parent.left;
}
if(parent.right != null){
if(pre != null){
pre.next = parent.right;
pre = pre.next;
}else
pre = parent.right;
}
parent = parent.next;
}
parent = next;
}
}
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
原题链接:https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
题目:前一题的继续,如果给定的是任意的二叉树,前题的代码还能用吗?
思路:显然,前一题不能用了。只好用层序遍历。
public void connect(TreeLinkNode root) {
if(root == null)
return;
TreeLinkNode parent = root,pre,next;
while(parent != null){
pre = null;
next = null;
while(parent != null){
if(next == null)
next = (parent.left == null) ? parent.right : parent.left;
if(parent.left != null){
if(pre != null){
pre.next = parent.left;
pre = pre.next;
}else
pre = parent.left;
}
if(parent.right != null){
if(pre != null){
pre.next = parent.right;
pre = pre.next;
}else
pre = parent.right;
}
parent = parent.next;
}
parent = next;
}
}
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