LeetCode-Populating Next Right Pointers in Each Node-优快云博客

本文深入解析了LeetCode中关于填充每个节点右侧指针的问题，提供了解法并附上C++代码实现。适用于理解树结构操作及递归算法的应用。

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作者：disappearedgod

文章出处：http://blog.youkuaiyun.com/disappearedgod/article/details/28238059

时间：2014-6-3

题目

Populating Next Right Pointers in Each Node

Total Accepted: 14951 Total Submissions: 43149

My Submissions

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.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */

解法

迭代解法

这道题参考了别人的算法

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null)
            return;
        TreeLinkNode t = new TreeLinkNode(root.val);
        if(root.left !=null){
            root.left.next = root.right;
        }
        if(root.right !=null && root.next!=null){
            root.right.next = root.next.left;
        }
        connect(root.left);
        connect(root.right);
    }
    
}

别人的一个C++代码

class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!root || !root->left)  return;
        
        auto parent = root, son = root->left;
        
        while (son)
        {
            auto back = son;
            
            while (son)
            {
                son->next = parent->right;
                son = son->next;
                parent = parent->next;
                son->next = parent?parent->left:nullptr;
                son = son->next;
            }
            
            parent = back;
            son = parent->left;
        }
    }
};

来源：http://blog.unieagle.net/2012/12/24/leetcode-problem-populating-next-right-pointers-in-each-node-level-traversal-of-binary-tree/