Populating Next Right Pointers in Each Node II LeetCode

Populating Next Right Pointers in Each Node II

Total Accepted: 18934 Total
Submissions: 62031My Submissions

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

Have you been asked this question in an interview?

/**
 * 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 ;
        }
        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
        queue.offer(root);
        while(!queue.isEmpty()){
            int qSize = queue.size();
            TreeLinkNode preNode = null;
            for (int i = 0; i < qSize; i++){
                TreeLinkNode curNode = queue.poll();
                curNode.next = preNode;
                preNode = curNode;
                if (curNode.right != null){
                    queue.offer(curNode.right);
                }
                if (curNode.left != null){
                    queue.offer(curNode.left);
                }
            }
        }
    }
}

时间: 2024-10-28 11:42:50

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描述:注意需要先self.connect(right)再self.connect(left),否则会有case通不过,原因是左边递归执行时依赖与右边的next已经建立,而先执行connect(left)的话右边还没有完成关系的建立. 代码: 1 class Solution: 2 # @param root, a tree node 3 # @return nothing 4 def doSth(self, nextNode, conNode): 5 while nextNode is not