Example 1:
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
Constraints:
-
The number of nodes in the given tree is less than
4096. -
-1000 <= node.val <= 1000
方法一：DFS，但递归会用到隐式非常量额外空间，不符题意。
方法二：BFS，但用到非常量额外空间的队列，不符题意。
方法三：子节点们利用父节点的next进行连接。
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
public class PopulatingNextRightPointersInEachNode {
public static class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {
}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
// 方法一：DFS
public Node connect(Node root) {
if (root == null)
return root;
connect(root, 0, new ArrayList<>());
return root;
}
private void connect(Node node, int level, List prevs) {
if (node == null)
return;
if (prevs.size() == level) {
prevs.add(node);
} else {
Node prev = prevs.get(level);
prev.next = node;
prevs.set(level, node);
}
connect(node.left, level + 1, prevs);
connect(node.right, level + 1, prevs);
}
// 方法二：BFS
public Node connect2(Node root) {
if (root == null)
return root;
LinkedList queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
Node prev = null;
for (int size = queue.size(); size > 0; size--) {
Node node = queue.poll();
if (prev != null) {
prev.next = node;
}
prev = node;
if (node.left != null)
queue.offer(node.left);
if (node.right != null)
queue.offer(node.right);
}
}
return root;
}
// 方法三：利用父节点的next
public Node connect3(Node root) {
if (root == null)
return null;
Node pre = root;
Node cur = null;
while (pre.left != null) {
cur = pre;
while (cur != null) {
if (cur.left != null)
cur.left.next = cur.right;
if (cur.right != null && cur.next != null)
cur.right.next = cur.next.left;
cur = cur.next;
}
pre = pre.left;
}
return root;
}
}
public class PopulatingNextRightPointersInEachNodeTest {
@Test
public void test() {
PopulatingNextRightPointersInEachNode obj = new PopulatingNextRightPointersInEachNode();
Node root = n(1, n(2, n(4), n(5)), n(3, n(6), n(7)));
root = obj.connect(root);
assertEquals("1,#,2,3,#,4,5,6,7,#", output(root));
}
@Test
public void test2() {
PopulatingNextRightPointersInEachNode obj = new PopulatingNextRightPointersInEachNode();
Node root = n(1, n(2, n(4), n(5)), n(3, n(6), n(7)));
root = obj.connect2(root);
assertEquals("1,#,2,3,#,4,5,6,7,#", output(root));
}
@Test
public void test3() {
PopulatingNextRightPointersInEachNode obj = new PopulatingNextRightPointersInEachNode();
Node root = n(1, n(2, n(4), n(5)), n(3, n(6), n(7)));
root = obj.connect3(root);
assertEquals("1,#,2,3,#,4,5,6,7,#", output(root));
}
private Node n(int v) {
return new Node(v);
}
private Node n(int v, Node l, Node r) {
return new Node(v , l, r, null);
}
private String output(Node root) {
if(root == null)
return "";
StringBuilder sb = new StringBuilder();
Node p = root;
while(p != null) {
Node p2 = p;
