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
对１１７题，要跳过空节点。 ３０％
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
TreeLinkNode prevLevelStart = root, prev, cur;
while (prevLevelStart != null) {
prev = prevLevelStart;
if (prev.left != null) {
cur = prev.left;
} else {
cur = prev.right;
prev = updatePrev(prev);
}
while (prev != null) {
if (cur == prev.left) {
if (prev.right != null) {
cur.next = prev.right;
cur = cur.next;
}
prev = updatePrev(prev);
} else {
if (prev.left != null) {
cur.next = prev.left;
cur = cur.next;
} else {
cur.next = prev.right;
cur = cur.next;
prev = updatePrev(prev);
}
}
}
prevLevelStart = updatePrevLevelStart(prevLevelStart);
}
}
public TreeLinkNode updatePrev(TreeLinkNode prev) {
prev = prev.next;
while (prev != null && prev.left == null && prev.right == null) {
prev = prev.next;
}
return prev;
}
public TreeLinkNode updatePrevLevelStart(TreeLinkNode prevLevelStart) {
while (prevLevelStart != null) {
if (prevLevelStart.left != null && (prevLevelStart.left.left != null || prevLevelStart.left.right != null)) {
return prevLevelStart.left;
} else if (prevLevelStart.right != null && (prevLevelStart.right.left != null || prevLevelStart.right.right != null)) {
return prevLevelStart.right;
} else {
prevLevelStart = prevLevelStart.next;
}
}
return null;
}
}
117. Populating Next Right Pointers in Each Node II
原文：http://www.cnblogs.com/yuchenkit/p/7192625.html