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.

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For example,
Given the following binary tree,

1/ \2 3/ \ \4 5 7

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After calling your function, the tree should look like:

1 -> NULL/ \2 -> 3 -> NULL/ \ \4-> 5 -> 7 -> NULL

?題目含義：從左到右連接每一層的節(jié)點(diǎn)

1 public void connect(TreeLinkNode root) { 2 while(root != null){ 3 TreeLinkNode tempChild = new TreeLinkNode(0); 4 TreeLinkNode currentChild = tempChild; 5 while(root!=null){ 6 if(root.left != null) { currentChild.next = root.left; currentChild = currentChild.next;} 7 if(root.right != null) { currentChild.next = root.right; currentChild = currentChild.next;} 8 root = root.next; 9 } 10 root = tempChild.next; 11 } 12 }

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轉(zhuǎn)載于:https://www.cnblogs.com/wzj4858/p/7715047.html

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