题目描述:

Given a binary tree

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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,

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     1
   /  \
  2    3
 / \  / \
4  5  6  7

After calling your function, the tree should look like:

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

给一颗完全二叉树的每一个节点确定它同层中的下一个节点. 规则比较简单:

  • 如果一个节点是父节点的左孩子, 那么它的next就是父节点的右孩子
  • 如果一个节点是父节点的右孩子, 那么分两种情况:
    • 父节点的next为null, 则该节点的next为null
    • 父节点的next不为null, 则该节点的next为父节点next节点的左孩子

为了知道一个节点的父节点, 在函数参数中要同时把节点的父节点传入.

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/**
 * 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, TreeLinkNode *parent = NULL) { // 与题目提供的函数原型略有不同S
        if(root == NULL)return;
        if(parent != NULL){
            if(root == parent->left){
                root->next = parent->right;
            }else if(root == parent->right){
                if(parent->next != NULL)root->next = parent->next->left;
                else root->next = NULL;
            }
        }
        connect(root->left, root);
        connect(root->right, root);
    }
};