Populating Next Right Pointers in Each Node II
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:You may only use constant extra space.For example,Given the following binary tree, 1 / \ 2 3 / \ \ 4 5 7After calling your function, the tree should look like: 1 -> NULL / \ 2 -> 3 -> NULL / \ \ 4-> 5 -> 7 -> NULLSOLUTION 1
本题还是可以用Level Traversal 轻松解出，连代码都可以跟上一个题目一模一样。
但是不符合空间复杂度的要求：constant extra space.
时间复杂度： O(N)
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) {12 return;13 }14 15 TreeLinkNode dummy = new TreeLinkNode(0);16 Queue q = new LinkedList ();17 18 q.offer(root);19 q.offer(dummy);20 21 while(!q.isEmpty()) {22 TreeLinkNode cur = q.poll();23 if (cur == dummy) {24 if (!q.isEmpty()) {25 q.offer(dummy);26 }27 continue;28 }29 30 if (q.peek() == dummy) {31 cur.next = null;32 } else {33 cur.next = q.peek(); 34 }35 36 if (cur.left != null) {37 q.offer(cur.left);38 }39 40 if (cur.right != null) {41 q.offer(cur.right);42 }43 }44 45 }46 }
SOLUTION 2
我们可以用递归解出。注意从右往左加next。否则的话 右边未建立，左边你没找不到next. Space Complexity:
时间复杂度： O(N)
1 public void connect(TreeLinkNode root) { 2 if (root == null) { 3 return; 4 } 5 6 TreeLinkNode cur = root.next; 7 TreeLinkNode next = null; 8 // this is very important. should exit after found the next. 9 while (cur != null && next == null) {10 if (cur.left != null) {11 next = cur.left;12 } else if (cur.right != null) {13 next = cur.right;14 } else {15 cur = cur.next;16 }17 }18 19 if (root.right != null) {20 root.right.next = next;21 next = root.right;22 }23 24 if (root.left != null) {25 root.left.next = next;26 }27 28 // The order is very important. We should deal with right first!29 connect(root.right);30 connect(root.left);31 } 2014.1229 redo:
但现在leetcode加强数据了，不管怎么优化，递归的版本再也不能通过，都TLE
1 // SOLUTION 2: REC 2 public void connect(TreeLinkNode root) { 3 if (root == null) { 4 return; 5 } 6 7 TreeLinkNode dummy = new TreeLinkNode(0); 8 TreeLinkNode pre = dummy; 9 10 if (root.left != null) {11 pre.next = root.left;12 pre = root.left;13 }14 15 if (root.right != null) {16 pre.right = root.right;17 pre = root.right;18 }19 20 if (root.left == null && root.right == null) {21 return;22 }23 24 // Try to find the next node;25 TreeLinkNode cur = root.next;26 TreeLinkNode next = null;27 while (cur != null) {28 if (cur.left != null) {29 next = cur.left;30 break;31 } else if (cur.right != null) {32 next = cur.right;33 break;34 } else {35 cur = cur.next;36 }37 }38 39 pre.next = next;40 41 if (root.right != null && (root.right.left != null || root.right.right != null)) {42 connect(root.right); 43 }44 45 if (root.left != null && (root.left.left != null || root.left.right != null)) {46 connect(root.left); 47 }48 49 }
SOLUTION 3
我们可以用Iterator 直接解出。并且不开辟额外的空间，也就是说空间复杂度是 O（1）
时间复杂度： O(N)
感谢 的作者
1 /* 2 Solution 3: iterator with O(1) space. 3 */ 4 public void connect(TreeLinkNode root) { 5 if (root == null) { 6 return; 7 } 8 9 connIterator(root);10 }11 12 /*13 This is a iterator version. 14 */15 public void connIterator(TreeLinkNode root) {16 TreeLinkNode leftEnd = root;17 while (leftEnd != null) {18 TreeLinkNode p = leftEnd;19 20 // Connect all the nodes in the next level together.21 while (p != null) {22 23 // find the 24 TreeLinkNode next = findLeftEnd(p.next);25 26 if (p.right != null) {27 p.right.next = next;28 next = p.right;29 }30 31 if (p.left != null) {32 p.left.next = next;33 }34 35 // continue to deal with the next point.36 p = p.next;37 }38 39 // Find the left end of the NEXT LEVEL.40 leftEnd = findLeftEnd(leftEnd);41 }42 43 }44 45 // Find out the left end of the next level of Root TreeNode.46 public TreeLinkNode findLeftEnd(TreeLinkNode root) {47 while (root != null) {48 if (root.left != null) {49 return root.left;50 }51 52 if (root.right != null) {53 return root.right;54 }55 56 root = root.next;57 }58 59 return null;60 } SOLUTION 4 (2014.1229):
在sol3基础上改进，引入dummynode，我们就不需要先找到最左边的点了。空间复杂度是 O（1）时间复杂度： O(N)
1 // SOLUTION 1: Iteration 2 public void connect1(TreeLinkNode root) { 3 if (root == null) { 4 return; 5 } 6 7 TreeLinkNode leftEnd = root; 8 9 // Bug 1: don't need " && leftEnd.left != null"10 while (leftEnd != null) {11 TreeLinkNode cur = leftEnd;12 13 TreeLinkNode dummy = new TreeLinkNode(0);14 TreeLinkNode pre = dummy;15 while (cur != null) {16 if (cur.left != null) {17 pre.next = cur.left;18 pre = cur.left;19 }20 21 if (cur.right != null) {22 pre.next = cur.right;23 pre = cur.right;24 }25 26 cur = cur.next;27 }28 leftEnd = dummy.next;29 }30 }
CODE ON GITHUB: