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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:
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
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//二叉树进行层次遍历 //判断最底层的条件:如果当前访问的层有任一节点存在子节点,说明当前层不是最底层。 class Solution{ public: TreeLinkNode* nextLeft; bool hasToTheEnd; void connect(TreeLinkNode *root) { if (!root) return; hasToTheEnd = false; int level = 1; while (!hasToTheEnd) { hasToTheEnd= true; nextLeft= NULL; VisitLevel(root, level); ++level; } } void VisitLevel(TreeLinkNode* node, int level) { if (level == 1) { if (nextLeft != NULL) { nextLeft->next = node; } nextLeft = node; if (node->left != NULL || node->right != NULL) { hasToTheEnd= false; } } if (node->left) VisitLevel(node->left, level - 1); if (node->right) VisitLevel(node->right, level - 1); } };
LeetCode 117:Populating Next Right Pointers in Each Node II
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原文地址:http://blog.csdn.net/geekmanong/article/details/50439036