题目:
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:
- 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
思路:
递归,然后把中间的节点给连起来
package tree; class TreeLinkNode {
int val;
TreeLinkNode left, right, next;
TreeLinkNode(int x) { val = x; }
} public class PopulatingNextRightPointersInEachNode { public void connect(TreeLinkNode root) {
if (root == null) return;
connect(root.left);
connect(root.right);
TreeLinkNode left = root.left;
TreeLinkNode right = root.right;
while (left != null && right != null) {
left.next = right;
left = left.right;
right = right.left;
}
} }