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:
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
分析：
把每层的Node放在list里面，然后遍历list的每个点，设置它们的nextRight.
1 /** 2 * Definition for binary tree with next pointer. 3 * public class TreeLinkNode { 4 * int val; 5 * TreeLinkNode left, right, next; 6 * TreeLinkNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public void connect(TreeLinkNode root) { 11 if (root == null) return; 12 ArrayList<TreeLinkNode> list1 = new ArrayList<TreeLinkNode>(); 13 ArrayList<TreeLinkNode> list2 = new ArrayList<TreeLinkNode>(); 14 15 list1.add(root); 16 17 while (!list1.isEmpty()) { 18 for (int i = 0; i < list1.size(); i++) { 19 if (list1.get(i).left != null) { 20 list2.add(list1.get(i).left); 21 } 22 if (list1.get(i).right != null) { 23 list2.add(list1.get(i).right); 24 } 25 } 26 for (int j = 0; j < list2.size() - 1; j++) { 27 list2.get(j).next = list2.get(j+1); 28 } 29 ArrayList<TreeLinkNode> temp = list1; 30 list1 = list2; 31 list2 = temp; 32 list2.clear(); 33 } 34 } 35 }
Another solution from Leetcode.
Here is a more elegant solution. The trick is to re-use the populated nextRight pointers. As mentioned earlier, we just need one more step for it to work. Before we passed the leftChild and rightChild to the recursion function itself, we connect the rightChild’s nextRight to the current node’s nextRight’s leftChild. In order for this to work, the current node’s nextRight pointer must be populated, which is true in this case.
1 void connect(Node* p) { 2 if (!p) return; 3 if (p->leftChild) 4 p->leftChild->nextRight = p->rightChild; 5 if (p->rightChild) 6 p->rightChild->nextRight = (p->nextRight) ? 7 p->nextRight->leftChild : 8 NULL; 9 connect(p->leftChild); 10 connect(p->rightChild); 11 }
Populating Next Right Pointers in Each Node
原文：http://www.cnblogs.com/beiyeqingteng/p/5731340.html