标签：tree space like 复杂度 following val pop binary nod
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
标签：tree space like 复杂度 following val pop binary nod
原文地址：http://www.cnblogs.com/yuchenkit/p/7192625.html
