116 Populating Next Right Pointers in Each Node

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本文介绍了一种在二叉树中填充每个节点的next指针的方法，使其指向右侧相邻节点的算法实现。针对完美二叉树及非完美二叉树的情况，提供了递归和迭代两种解决方案。

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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,

After calling your function, the tree should look like:

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

Follow up, what if the tree is not prefect binary tree.

这道题的follow up 中 binary tree不一定是完整的所以root.right.next指向是不定的

思路就是先找到root.right 右边第一个可行node存储注意这道题应该先遍历右边然后再左边

递归:

/**
 * 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) {
        if (root == null)
            return;
        TreeLinkNode dummy = new TreeLinkNode(0);
        
        for (TreeLinkNode curr = root, prev = dummy; curr != null;
            curr = curr.next) {
            if (curr.left != null) {
                prev.next = curr.left;
                prev = prev.next;
            }
            
            if (curr.right != null) {
                prev.next = curr.right;
                prev = prev.next;
            }
        }
        connect(dummy.next);
    }
}

迭代:

/**
 * 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) {
        while (root != null) {
            TreeLinkNode next = null;
            TreeLinkNode prev = null;
            
            for (; root != null; root = root.next) {
                if (next == null)
                    next = root.left != null ? root.left : root.right;
                if (root.left != null) {
                    if (prev != null)
                        prev.next = root.left;
                    prev = root.left;
                }
                
                if (root.right != null) {
                    if (prev != null)
                        prev.next = root.right;
                    prev = root.right;
                }
            }
            root = next;
        }
    }
}