LeetCode-Populating Next Right Pointers in Each Node-填充结点的右指针-二叉树递归

https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

构造一个按深度索引的数组，存储的当前遍历的结点在这一层的最右边的结点。

由前序遍历的特点可知同一层左边的儿子先被访问，过一段时间后会访问右边的儿子。利用这个特点连接next数组即可。

需要注意vec.size()-1在vec=0时不是-1，而是一个很大的整数（溢出了）。

/*** 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) {}* };*/typedef TreeLinkNode scnode;
class Solution {
public:vector<scnode *> ps;void PreOrder(scnode *rt,int l){if (rt==NULL) return;if (l+1>ps.size()){ps.push_back(NULL);}if (ps[l]!=NULL) ps[l]->next=rt;ps[l]=rt;rt->next=NULL;PreOrder(rt->left,l+1);PreOrder(rt->right,l+1);}void connect(TreeLinkNode *root) {PreOrder(root,0);}
};

转载于:https://www.cnblogs.com/yangsc/p/4025150.html

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