【leetcode】Populating Next Right Pointers in Each Node

Populating Next Right Pointers in Each Node

链接：https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

描述：

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

解法：

本题比较简单，注意 判断NULL的情况。

代码如下：

    void connect(TreeLinkNode *root) {
    	if( root == NULL) return ;
    	TreeLinkNode *preP = root;
    	TreeLinkNode *curP = root->left;
    	root->next = NULL;
    
    	while( curP )
    	{
    		TreeLinkNode *tempP = preP;
    		while( curP )
    		{
    			if( preP == NULL){
    				curP->next = NULL;
    				curP = NULL;
    			}else if( curP == preP->left){
    				curP->next = preP->right;
    				curP = curP->next;
    				preP = preP->next;
    			}else{
    				curP->next = preP->left;
    				curP = curP->next;
    			}
    		}
    		preP = tempP->left;
    		curP = preP->left;
    	}
    }

Populating Next Right Pointers in Each Node II

链接：https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

描述：

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

解法：

本题比上一题，复杂，非递归解法如下：

    void connect(TreeLinkNode *root) {
    	if(root == NULL) return ;
    	root->next = NULL;
    	TreeLinkNode *prep = root;
    	TreeLinkNode *curp;
    	if(root->left)
    		curp = root->left;
    	else
    		curp = root->right;
    
    	while(curp)
    	{
    		TreeLinkNode *tempcurp=curp;
    		while(curp)
    		{
    			if(prep == NULL)
    			{
    				curp->next = NULL;
    				break;
    			}else if( curp == prep->right)
    			{
    				prep = prep->next;
    			}else if( curp == prep->left)
    			{
    				if( prep->right )
    				{
    					curp->next = prep->right;
    					curp = curp->next;
    				}
    				prep = prep->next;
    			}else{
    				while(prep && prep->left == NULL && prep->right == NULL)
    					prep = prep->next;
    				if(prep){
    					if(prep->left)
    						curp->next  = prep->left;
    					else
    						curp->next = prep->right;
    					curp = curp->next;
    				}
    			}
    		}
    		prep = tempcurp;
    		while(prep && prep->left == NULL && prep->right == NULL)
    			prep = prep->next;
    		if(prep && prep->left)
    			curp = prep->left;
    		else if(prep && prep->right)
    			curp = prep->right;
    		else
    			break;
    	}
    }

递归解法：

    void connect(TreeLinkNode *root) {
        if(root == NULL) return;
        TreeLinkNode dumy(-1);
        //dumy.next = root;
        
        for(TreeLinkNode *cur = root, *pre = &dumy; cur ; cur = cur->next){
            if(cur->left != NULL){
                pre->next = cur->left;
                pre = pre->next;
            }
            if(cur->right != NULL){
                pre->next = cur->right;
                pre = pre->next;
            }
        }
        connect(dumy.next);
    }