117-Populating Next Right Pointers in Each Node II

Description

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.

---

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

---

問題描述

給定二叉樹的節點結構如下

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *next;
}

設定next指標指向右邊的節點， 如果不存在右邊的節點， 將next設定為空

(這題與116的區別是， 樹不一定是滿二叉樹)

---

問題分析

解法1為這種型別的題目經常用到的一種解法， 關鍵是利用好height

解法2將二叉樹的每一層級當作連結串列給連線起來

---

解法1

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null) return;

        List<TreeLinkNode> res = new ArrayList();
        connectFunc(root, res, 0);
    }
    public void connectFunc(TreeLinkNode root, List<TreeLinkNode> res, int height){
        if(root == null) return;

        if(height == res.size()) res.add(root);
        else{
            res.get(height).next = root;
            res.set(height,root);
        }

        connectFunc(root.left, res, height + 1);
        connectFunc(root.right, res, height + 1);
    }
}

---

解法2

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null) return;

        TreeLinkNode dummyHead = new TreeLinkNode(0);
        TreeLinkNode travel = dummyHead;

        while(root != null){
            if(root.left != null){
                travel.next = root.left;
                travel = travel.next;
            }
            if(root.right != null){
                travel.next = root.right;
                travel = travel.next;
            }
            root = root.next;
        }

        connect(dummyHead.next);
    }
}