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leetcode--Populating Next Right Pointers in Each Node II

2024-07-07 03:45:45   221人阅读

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

?
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public class Solution {
    public void connect(TreeLinkNode root) {
        if(root != null){
            root.next = null;
            TreeLinkNode topLevel = root;
            TreeLinkNode nextLevelHead = null;
            TreeLinkNode node = null;
            while(topLevel != null){
                //looking for the head of next level
                while(topLevel != null && nextLevelHead == null){
                    if(topLevel.left != null){
                        nextLevelHead = topLevel.left;
                        node = nextLevelHead;
                    }
                    if(topLevel.right != null){
                        if(node == null){
                            nextLevelHead = topLevel.right;
                            node = nextLevelHead;
                        }
                        else{
                            node.next = topLevel.right;
                            node = node.next;
                        }
                    }
                     
                    topLevel = topLevel.next;
                }
                 
                while(topLevel != null){
                    if(topLevel.left != null){
                        node.next = topLevel.left;
                        node = node.next;
                    }
                    if(topLevel.right != null){
                        node.next = topLevel.right;
                        node = node.next;
                    }
                    topLevel = topLevel.next;
                }
                 
                if(node != null){
                    node.next = null;
                    node = node.next;
                }
                topLevel = nextLevelHead;
                nextLevelHead = null;
            }
        }
    }
}

　　


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