leetcode 117. Populating Next Right Pointers in Each Node II

最新推荐文章于 2022-06-02 20:32:40 发布

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最新推荐文章于 2022-06-02 20:32:40 发布

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分类专栏： leetcode 文章标签： leetcode Tree

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针对Populating Next Right Pointers in Each Node问题的进阶版本，探讨如何在任意二叉树中填充节点间的next指针，确保常数额外空间使用。通过调整遍历顺序，先处理右子树再处理左子树，实现高效解决方案。

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Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.
For 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
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如果按照上一题的解法，那就需要去寻找next的位置
这里需要注意的是先做右子树，再做左子树
比较清晰易懂的解法

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root==null) return;
        TreeLinkNode right =null ;
        if(root.left!=null&&root.right!=null){
            root.left.next = root.right;
            root.right.next = getNext(root.next);
        }else{
            if(root.left!=null) right = root.left;
            if(root.right!=null) right = root.right;
            if(right!=null) right.next = getNext(root.next);
        }
        // connect(root.next);
        connect(root.right);
        connect(root.left);
    }

    TreeLinkNode getNext(TreeLinkNode p){
        if(p==null) return null;
        if(p.left!=null){
            return p.left;
        }else if(p.right!=null){
            return p.right;
        }else{
            return getNext(p.next);
        }
    }
}