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LeetCode 117 - Populating Next Right Pointers in Each Node II

 
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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,

         1
       /  \
      2    3
     / \    \
    4   5    7

 

After calling your function, the tree should look like:

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

Solution 1:

递归版本。

public void connect(TreeLinkNode root) {
    if(root == null) return;
    TreeLinkNode p = root.next;
    while(p!=null) {
        if(p.left != null) {
            p = p.left; 
            break;
        }
        if(p.right != null) {
            p = p.right; 
            break;
        }
        p = p.next;
    }
    if(root.right != null) {
        root.right.next = p;
        p = root.right;
    } 
    if(root.left != null) {
        root.left.next = p;
    }
    connect(root.right);
    connect(root.left);
}

 

Solution 2:

非递归版本。

public void connect(TreeLinkNode root)  {
    if(root == null)  return;  
    TreeLinkNode lastHead = root;  
    TreeLinkNode curHead = null, curPrev = null;
    while(lastHead!=null)  {  
        while(lastHead != null) {
            if(lastHead.left!=null) {  
                if(curHead == null) {  
                    curHead = lastHead.left;  
                    curPrev = curHead;  
                } else {  
                    curPrev.next = lastHead.left;  
                    curPrev = curPrev.next;  
                }  
            }  
            if(lastHead.right!=null) {  
                if(curHead == null) {  
                    curHead = lastHead.right;  
                    curPrev = curHead;  
                } else {  
                    curPrev.next = lastHead.right;  
                    curPrev = curPrev.next;  
                }  
            }                  
            lastHead = lastHead.next; 
        }
        lastHead = curHead;  
        curHead = null;  
    }  
}

 

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