leetcode 79: Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-09-24 12:05:40 发布

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最新推荐文章于 2021-09-24 12:05:40 发布

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本文探讨了在不使用额外空间的情况下，如何为给定的任意二叉树节点实现连接操作，使得每个节点能够访问其下一个右侧节点。通过实例演示了解决方案，并提供了Java和C/C++两种语言的实现方式。

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Populating Next Right Pointers in Each Node IIOct 28 '12

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

/**
 * 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) {
        // Start typing your Java solution below
        // DO NOT write main() function
        
        if(root==null) return;
        
        Queue<TreeLinkNode> q1 = new LinkedList<TreeLinkNode>();
        Queue<TreeLinkNode> q2 = new LinkedList<TreeLinkNode>();
        
        q1.offer( root );
        root.next = null;
        while( !q1.isEmpty() ) {
            TreeLinkNode t = q1.poll();
            if(t.left != null) q2.offer( t.left);
            if(t.right!=null) q2.offer( t.right );
            
            t.next = q1.isEmpty() ? null : q1.peek();
            
            if(q1.isEmpty()) {
                Queue<TreeLinkNode> temp = q1;
                q1 = q2;
                q2 = temp;
            }
        }
    }
}

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if(root==nullptr) return;
        
        queue<TreeLinkNode*> que;
        que.push( root);
        
        int q1 = 1;
        int q2 = 0;
        while(!que.empty() ) {
            TreeLinkNode * n = que.front();
            que.pop();
            --q1;
            
            if(n->left!=nullptr) {
                que.push(n->left);
                ++q2;
            }
            if(n->right!=nullptr){
                que.push(n->right);
                ++q2;
            }
            
            if(q1==0) {
                swap(q1,q2);
            } else {
                n->next = que.front();
            }
        }
    }
};