## [LeetCode] Populating Next Right Pointers in Each Node I, II

Given a binary tree

    struct TreeLinkNode {
}


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

/**
* Definition for binary tree with next pointer.
*  int val;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
if(NULL == root) return;
while(root -> left != NULL){
curLev = root;
while(curLev != NULL){
curLev -> left -> next = curLev -> right;
if(curLev -> next != NULL)
curLev -> right -> next = curLev -> next -> left;
curLev = curLev -> next;
}
root = root -> left;
}
}
};

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

class Solution {
public:
if(NULL == root) return;
while(root != NULL){
start = findStartNodeNextLev(root);
curNode = start;
nextNode = findNextNodeNextLev(root, start);
while(nextNode != NULL){
curNode -> next = nextNode;
curNode = nextNode;
nextNode = findNextNodeNextLev(root, curNode);
}
root = start;
}
}
private:
if(cur -> left == curNextLev && cur -> right != NULL){
return cur -> right;
}else{
while(cur -> next != NULL){
cur = cur -> next;
if(cur -> left != NULL && cur -> left != curNextLev) return cur -> left;
if(cur -> right != NULL && cur -> right != curNextLev) return cur -> right;
}
}
return NULL;
}

if(NULL == node) return NULL;
if(node -> left != NULL) return node -> left;
return findNextNodeNextLev(node, node -> left);
}
};

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Felix原创，转载请注明出处，感谢博客园！

posted on 2014-04-06 05:37  Felix Fang  阅读(6375)  评论(0编辑  收藏  举报