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
public void connect(TreeLinkNode root) {
if (null == root) {
return;
}
TreeLinkNode cur = root.next;
TreeLinkNode p = null;
while (cur != null) { // find last right node (left or right)
if (cur.left != null) {
p = cur.left;
break;
}
if (cur.right != null) {
p = cur.right;
break;
}
cur = cur.next;
}
if (root.right != null) {
root.right.next = p;
}
if (root.left != null) {
root.left.next = root.right != null ? root.right : p;
}
connect(root.right); // from right to left
connect(root.left);
} public void connect(TreeLinkNode root) {
if (null == root) {
return;
}
LinkedList<TreeLinkNode> Q = new LinkedList<TreeLinkNode>(); // save one
// line
// root(s)
LinkedList<TreeLinkNode> Q2 = new LinkedList<TreeLinkNode>();
; // save next one line root(s), swap with Q
Q.push(root);
while (!Q.isEmpty()) {
TreeLinkNode tmp = Q.getFirst();
Q.pop();
if (tmp.left != null)
Q2.add(tmp.left);
if (tmp.right != null)
Q2.add(tmp.right);
if (Q.isEmpty()) {
tmp.next = null;
LinkedList<TreeLinkNode> tmpQ = Q; // swap queue
Q = Q2;
Q2 = tmpQ;
} else {
tmp.next = Q.getFirst();
}
}
}注意点: public void connect(TreeLinkNode root) {
LinkedList<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
if (root == null)
return;
TreeLinkNode p;
queue.add(root);
queue.add(null);// flag
while (!queue.isEmpty()) {
p = queue.pop();
if (p != null) {
if (p.left != null) {
queue.add(p.left);
}
if (p.right != null) {
queue.add(p.right);
}
p.next = queue.getFirst();
} else {
if (queue.isEmpty()) {
return;
}
queue.add(null);
}
}
}LeetCode 117 Populating Next Right Pointers in Each Node II
原文地址:http://blog.csdn.net/mlweixiao/article/details/40649871