[LeetCode73]Populating Next Right Pointers in Each Node II

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

Analysis:

与上一题类似，唯一的不同是每次要先找到一个第一个有效的next链接节点，并且递归的时候要先处理右子树，再处理左子树.

Java

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;
        if(root.left!=null)
        	root.left.next = root.right != null ? root.right:p;
        connect(root.right);
        connect(root.left);
    }

c++

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;
     }
     if(root->left != NULL){
        root->left->next = root->right ? root->right : p;
     }
     connect(root->right);
     connect(root->left);
    }