[leetcode]48 Binary Tree Right Side View
2015-04-07 13:32
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题目链接:https://leetcode.com/problems/binary-tree-right-side-view/
Runtimes:9ms
For example:
Given the following binary tree,
1 <—
/ \
2 3 <—
\ \
5 4 <—
You should return [1, 3, 4].
Runtimes:9ms
1、问题
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.For example:
Given the following binary tree,
1 <—
/ \
2 3 <—
\ \
5 4 <—
You should return [1, 3, 4].
2、分析
说到底,就是求一棵树的右侧面照,即每一层从右往左第一个数字。肯定是分层遍历,但是这个过程有优化的地方,以前做类似分层遍历题目的时候都是多开几个大vector存储数据,现在又有了新的辨别方式,剩下了很多空间以及时间。3、小结
只要记录一层最后访问的那个节点即可,即preh < h时,pret就是要求的最后节点。用队列会和遍历的顺序一样,选择正确的数据结构也是很重要的。4、实现
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> rightSideView(TreeNode *root) { queue <TreeNode *> tv; queue <int> iv; vector<int> rv; if(NULL == root) return rv; tv.push(root); iv.push(1); int preh = INT_MAX; TreeNode *pret; while(!tv.empty()) { TreeNode * t = tv.front(); tv.pop(); int h = iv.front(); iv.pop(); if(preh < h) rv.push_back(pret->val); if(NULL != t->left) { tv.push(t->left); iv.push(h + 1); } if(NULL != t->right) { tv.push(t->right); iv.push(h + 1); } preh = h; pret = t; } rv.push_back(pret->val); return rv; } };
5、反思
从编程之美吸收了很多优化方法,一本不错的书。相关文章推荐
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