LeetCode 117. Populating Next Right Pointers in Each Node II（右邻）

原题网址：https://leetcode.com/problems/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    7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \    \
4-> 5 -> 7 -> NULL

方法一：对于每个节点，寻找所有的深度的子节点，直到找完全部。

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeLinkNode leftmost(TreeLinkNode root, int depth) {
if (root == null || depth == 0) return root;
TreeLinkNode leftmost = leftmost(root.left, depth-1);
if (leftmost != null) return leftmost;
leftmost = leftmost(root.right, depth-1);
return leftmost;
}
private TreeLinkNode rightmost(TreeLinkNode root, int depth) {
if (root == null || depth == 0) return root;
TreeLinkNode rightmost = rightmost(root.right, depth-1);
if (rightmost != null) return rightmost;
rightmost = rightmost(root.left, depth-1);
return rightmost;
}
public void connect(TreeLinkNode root) {
if (root == null) return;
if (root.left != null) connect(root.left);
if (root.right != null) connect(root.right);
for(int i=1;; i++) {
TreeLinkNode left = rightmost(root.left, i-1);
if (left == null) break;
TreeLinkNode right = leftmost(root.right, i-1);
if (right == null) break;
left.next = right;
}
}
}

方法二：利用子问题，层层连接。

/**
* 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 upper = root;
TreeLinkNode leftmost = null;
TreeLinkNode bottom = null;
while (upper != null) {
while (upper != null) {
if (upper.left != null && bottom != upper.left && (upper.right == null || bottom != upper.right)) {
if (leftmost == null) {
leftmost = upper.left;
bottom = leftmost;
} else {
bottom.next = upper.left;
bottom = bottom.next;
}
} else if (upper.right != null && bottom != upper.right) {
if (leftmost == null) {
leftmost = upper.right;
bottom = leftmost;
} else {
bottom.next = upper.right;
bottom = bottom.next;
}
} else {
upper = upper.next;
}
}
upper = leftmost;
leftmost = null;
bottom = null;
}
}
}