[LeetCode] Populating Next Right Pointers in Each Node II
2013-11-01 10:25
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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,
After calling your function, the tree should look like:
虽然代码很长,但是逻辑还是很清楚的。
当遍历到一层时,如果有左孩子,没有右孩子,那么左孩子的next就指向父节点的下一个有孩子节点的第一个孩子;如果既有左孩子,又有右孩子,那么左孩子的next指向有孩子,右孩子的next指向父节点的下一个有孩子节点的第一个孩子;如果有右孩子,没有左孩子,那么右孩子的next指向父节点的下一个有孩子节点的第一个孩子,这种情况和第二种情况有重叠;当然,如果没有孩子节点,那么就遍历到下一个节点。
不过上面的代码有些冗余,在两个if中代码相似度很高,可以将上面的两个if简化:
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 问题描述:给定一个二叉树,使每个节点的next指针指向它的右边的节点,和之前的Populating Next Right Pointers in Each Node类似,只是这次的二叉树不是完全二叉树,但是方法和思想与之前的一样。 用一个指针遍历每层的最左节点,用另一个指针遍历前面的指针所在层的所有节点,由于不是完全二叉树,这里引入了另外一个函数find_has_child(),它返回从参数节点开始,该层第一个有孩子的节点。
class Solution { public: TreeLinkNode *find_has_child(TreeLinkNode *parent) { TreeLinkNode *p = parent; while(p && !(p->left) && !(p->right)) p = p->next; return p; } void connect(TreeLinkNode *root) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. TreeLinkNode *parent = root; if(!parent) return; parent->next = NULL; while(parent) { TreeLinkNode *level_parent = parent; while(level_parent->next) { TreeLinkNode *lchild = level_parent->left; TreeLinkNode *rchild = level_parent->right; if(lchild && !rchild) { TreeLinkNode *p = find_has_child(level_parent->next); if(p) { if(p->left) { lchild->next = p->left; } else { lchild->next = p->right; } level_parent = p; continue; } else { lchild->next = NULL; break; } } if(rchild) { if(lchild) lchild->next = rchild; TreeLinkNode *p = find_has_child(level_parent->next); if(p) { if(p->left) { rchild->next = p->left; } else { rchild->next = p->right; } level_parent = p; continue; } else { rchild->next = NULL; break; } } level_parent = level_parent->next; } if(level_parent->left) level_parent->left->next = level_parent->right; else if(level_parent->right) level_parent->right->next = NULL; if(parent->left) { parent = parent->left; } else if(parent->right) { parent = parent->right; } else { TreeLinkNode *p = find_has_child(parent->next); if(p) { if(p->left) { parent = p->left; } else if(p->right) { parent = p->right; } } else { parent = NULL; } } } } };
虽然代码很长,但是逻辑还是很清楚的。
当遍历到一层时,如果有左孩子,没有右孩子,那么左孩子的next就指向父节点的下一个有孩子节点的第一个孩子;如果既有左孩子,又有右孩子,那么左孩子的next指向有孩子,右孩子的next指向父节点的下一个有孩子节点的第一个孩子;如果有右孩子,没有左孩子,那么右孩子的next指向父节点的下一个有孩子节点的第一个孩子,这种情况和第二种情况有重叠;当然,如果没有孩子节点,那么就遍历到下一个节点。
不过上面的代码有些冗余,在两个if中代码相似度很高,可以将上面的两个if简化:
if(lchild && rchild) lchild->next = rchild; if(lchild || rchild) { TreeLinkNode *p = find_has_child(level_parent->next); TreeLinkNode *q = NULL; if(lchild && !rchild) q = lchild; if(rchild) q = rchild; if(p) { if(p->left) q->next = p->left; else q->next = p->right; level_parent = p; continue; } else { q->next = NULL; break; } }
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