116 Populating Next Right Pointers in Each Node **** 重要的题目

Given a binary tree

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *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.

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
Example:

Given the following perfect binary tree,

     1
   /  \
  2    3
 / \  / \
4  5  6  7
After calling your function, the tree should look like:

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

• [x] ### 重要的是traversal order的选择
• [x] ### recursion
• [x] ### O（n） time and space

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
//妥妥的先序遍历
//因为你要先把 2 和 3 链接起来
public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            root.left.next = root.right;
        }
        if (root.next != null && root.right != null) {
            root.right.next = root.next.left;
        }
        connect(root.left);
        connect(root.right);
    }
}

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    //time o(n) space O(1)
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return;
        }
        TreeLinkNode start = root;
        while (start != null) {
            TreeLinkNode curt = start;
            while (curt != null) {
            if (curt.left != null) {
                curt.left.next = curt.right;
            }
            if (curt.next != null && curt.right != null) {
                curt.right.next = curt.next.left;
            }
            curt = curt.next;
            }
            start = start.left;
        }
    }
}