117. Populating Next Right Pointers in Each Node II

摘要： Given a binary tree struct Node { int val; Node *left; Node *right; Node *next; } Populate each next pointer to poin...

Given a binary tree

struct Node {

int val;

Node *left;

Node *right;

Node *next;

}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

1. You may only use constant extra space.
2. Recursive approach is fine, implicit stack space does not count as extra space for this problem.

難度：medium

題目：給定二叉樹，計算結點指向其相鄰右結點的下一跳指標。如果沒有相鄰的右結點則next為NULL.

思路：遞迴

Runtime: 3 ms, faster than 18.40% of Java online submissions for Populating Next Right Pointers in Each Node II.

Memory Usage: 51.8 MB, less than 100.00% of Java online submissions for Populating Next Right Pointers in Each Node II.

/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;

public Node() {}

public Node(int _val,Node _left,Node _right,Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
Node ptr = root;
while (ptr != null) {
ptr = buildNodeNext(ptr);
}

return root;
}

private Node buildNodeNext(Node head) {
if (null == head) {
return null;
}
Node nextLevelLeftMost = buildNodeNext(head.next);
Node left = head.left;
Node right = head.right;
if (left != null && right != null) {
left.next = right;
right.next = nextLevelLeftMost;
return left;
}

if (left != null && right == null) {
left.next = nextLevelLeftMost;
return left;
} 

if (left == null && right != null) {
right.next = nextLevelLeftMost;
return right;
}

return nextLevelLeftMost;
}
}