leetCode 116.Populating Next Right Pointers in Each Node （替节点填充右指针） 解题思路和方法

leetCode 116.Populating Next Right Pointers in Each Node （为节点填充右指针） 解题思路和方法

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.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For 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

思路：为二叉树的每个节点均添加右指针，指向下一个水平序节点。其思路是根据上一层的右指针来求得下一个水平序节点。

具体代码如下：

/**
 * 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) {
        dfs(root);
    }
    
    /**
     * dfs，通过next来求解，假设上一层next指针均设置好，
     * 则下一层的next指针也可依次设置
     * 因为是完美二叉树，所有的底层左子树均存在，不存在的说明结束
     */ 
    private void dfs(TreeLinkNode root){
        if(root == null || root.left == null){
            return;
        }
        TreeLinkNode p = root;
        
        while(root != null){
        	root.left.next = root.right;
        	if(root.next != null){
        		root.right.next = root.next.left;
        	}
        	root = root.next;
        }
        dfs(p.left);
    }
}