117. Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-01-29 16:34:55 发布

SnailRunFast

最新推荐文章于 2021-01-29 16:34:55 发布

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分类专栏： Leet Code 文章标签： Populating Next Righ

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本文探讨了如何在不使用额外空间的情况下，为二叉树的每个节点填充指向其右侧相邻节点的指针。通过递归从左到右、从上到下的方式实现，确保了空间复杂度为常数。

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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,

After calling your function, the tree should look like:

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

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Solution:

Tips:

recursion in from left to right, up to down.

Java Code:

/**
 * 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) {
        treeRecursion(root);
    }
    
    void listRecursion(TreeLinkNode root, TreeLinkNode[] lastNode){
        if(null == root) {
            return;
        }
        
        listRecursion(root.next, lastNode);
        if(root.right != null){
            root.right.next = lastNode[0];
            lastNode[0] = root.right;
        }
        if(root.left != null){
            root.left.next = lastNode[0];
            lastNode[0] = root.left;
        }
    }
    
    void treeRecursion(TreeLinkNode root){
        if(null == root) {
            return;
        }
        
        TreeLinkNode[] lastNode = {null};
        listRecursion(root, lastNode);
        treeRecursion(lastNode[0]);
    }
}