LeetCode Binary Tree Right Side View 树的层次遍历
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思路:
每层的最后一个元素,即使用层次遍历一遍。
时间复杂度O(N),空间复杂度O(N)。N为节点个数。
/** * Definition for a binary tree node. * 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) { vector<int> ans; if(root == NULL) return ans; queue<TreeNode*> q; q.push(root); TreeNode *node; while(!q.empty()) { int size = q.size(); for(int i = 0; i < size; ++i) { node = q.front(); q.pop(); if(node->left) { q.push(node->left); } if(node->right) { q.push(node->right); } } ans.push_back(node->val); } return ans; }};
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