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populateNextPointer.java
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81 lines (80 loc) · 2.58 KB
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import java.util.*;
class TreeLinkNode {
int val;
TreeLinkNode left, right, next;
TreeLinkNode(int x) { val = x; }
}
public class populateNextPointer {
/* Approach 1: 对空间复杂度没有要求
就是很基本的用queue 做 level order traverse
* */
public void connect(TreeLinkNode root) {
if (root == null) {
return;
}
Queue<TreeLinkNode> q = new LinkedList<>();
q.add(root);
while (!q.isEmpty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
TreeLinkNode cur = q.remove();
if (i == size - 1) {
cur.next = null;
} else {
cur.next = q.peek();
}
if (cur.left != null) {
q.add(cur.left);
}
if (cur.right != null) {
q.add(cur.right);
}
}
}
}
/* Approach 2:
Considering in previous levels, neighbors are connected by "next". So we don't need to use
queue to do level order traverse. We can make use of queue.
Two while loops:
一个是一层一层往下走
一个是在当前层从左往右走
因为tree 不是perfect的，所以要记录上下一层的 startNode -> levelHead
* */
public void connect2(TreeLinkNode root) {
if (root == null) {
return;
}
TreeLinkNode levelHead = root;
TreeLinkNode prev;
TreeLinkNode cur;
// boolean findHead = false; don't need that, we can see wheter prev ? null
// 不要动root
while (levelHead != null) {
cur = levelHead;
levelHead = null;
prev = null;
// traverse the level
while (cur != null) {
if (cur.left != null) {
if (prev == null) {// find the head of the level
levelHead = cur.left;
} else {
prev.next = cur.left;
}
prev = cur.left;
}
if (cur.right != null) {
if (prev == null) {// find the head of next level
levelHead = cur.right;
} else {
prev.next = cur.right;
}
prev = cur.right;
}
cur = cur.next;
}
//go down to next level
//root = levelHead;
}
}
}