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PopulatingNextRightPointersInEachNodeII.java
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PopulatingNextRightPointersInEachNodeII.java
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/**
* Practiced on 12/18/2012
*
* 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
*
* =======================================
* First wrote a tedious code, and then revised to the following code.
*/
package com.congli.leetcode;
public class PopulatingNextRightPointersInEachNodeII {
public void connect(TreeLinkNode root)
{
if(root == null) return;
TreeLinkNode parentStart = root;
TreeLinkNode childStart = root.left == null ? root.right : root.left;
TreeLinkNode parent = parentStart;
TreeLinkNode child = childStart;
while(childStart != null)
{
while(parent != null)
{
// Find the next node for child.
TreeLinkNode nextChild = null;
if(child == parent.left && parent.right != null)
nextChild = parent.right;
else
{
parent = parent.next;
while(parent != null && parent.left == null && parent.right == null)
parent = parent.next;
if(parent != null)
nextChild = parent.left != null ? parent.left : parent.right;
}
child.next = nextChild;
child = child.next;
}
parentStart = childStart;
childStart = findChildStart(childStart);
parent = parentStart;
child = childStart;
}
}
private TreeLinkNode findChildStart(TreeLinkNode childStart)
{
while(childStart != null && childStart.left == null && childStart.right == null)
childStart = childStart.next;
if(childStart == null) return null;
return childStart.left != null ? childStart.left : childStart.right;
}
}