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[LeetCode]Binary Tree Level Order Traversal
【题目】
Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / 9 20 / 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
【代码】
/********************************* * 日期:2014-12-08 * 作者:SJF0115 * 题号: Binary Tree Level Order Traversal * 来源:https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ * 结果:AC * 来源:LeetCode * 总结: **********************************/ #include <iostream> #include <malloc.h> #include <vector> #include <queue> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: vector<vector<int> > levelOrder(TreeNode *root) { vector<int> level; vector<vector<int> > levels; if(root == NULL){ return levels; } //入队列 queue<TreeNode*> queue; queue.push(root); //当前层节点个数 int count = 1; //下一层节点个数 int nextCount = 0; // 层次遍历 while(!queue.empty()){ //取队列头元素 TreeNode *p = queue.front(); //入队列 level.push_back(p->val); queue.pop(); count--; //左子树 if(p->left != NULL){ queue.push(p->left); // 下一层节点数加一 nextCount++; }//if // 右子树 if(p->right != NULL){ queue.push(p->right); // 下一层节点数加一 nextCount++; }//if // 当前层访问完毕 if(count == 0){ count = nextCount; nextCount = 0; levels.push_back(level); level.clear(); }//if }//while return levels; } }; //按先序序列创建二叉树 int CreateBTree(TreeNode* &T){ char data; //按先序次序输入二叉树中结点的值(一个字符),‘#’表示空树 cin>>data; if(data =http://www.mamicode.com/= '#'){>[LeetCode]Binary Tree Level Order Traversal
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