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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
指出每个节点的下一个右侧节点，前一篇博客的那个思路也可以继续使用，但由于这里的二叉树是任意的，所以应该用函数找到下一层的开始节点以及下一层的下一个节点，代码如下：
1 class Solution{ 2 public: 3 void connect(TreeLinkNode *root) 4 { 5 TreeLinkNode * prev, * curr, * start; 6 if(!root) return; 7 while(root){ 8 start = findNextLevelStartNode(root); 9 prev = start; 10 curr = findNextLevelNextNode(root, prev); 11 while(curr){ 12 prev->next = curr; 13 prev = curr; 14 curr = findNextLevelNextNode(root, curr); 15 } 16 root = start; 17 } 18 } 19 private: 20 TreeLinkNode * findNextLevelNextNode(TreeLinkNode * & node, TreeLinkNode * curr)//注意使用引用 21 { 22 if(node->left == curr && node->right) 23 return node->right; 24 else{ 25 while(node->next){ 26 node = node->next; 27 if(node->left != NULL && node->left != curr) return node->left; 28 if(node->right != NULL && node->right != curr) return node->right; 29 } 30 } 31 return NULL; 32 } 33 34 TreeLinkNode * findNextLevelStartNode(TreeLinkNode * node) 35 { 36 if(!node) return NULL; 37 if(node->left) 38 return node->left; 39 else return findNextLevelNextNode(node, node->left); 40 } 41 };
LeetCode OJ：Populating Next Right Pointers in Each Node II（指出每一个节点的下一个右侧节点II）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4915789.html
