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leetcode-Binary Tree Level Order Traversal 二叉树层序遍历
leetcode-Binary Tree Level Order Traversal 二叉树层序遍历
#include<stdio.h> #include<queue> using namespace std; typedef struct BiTree { int val; struct BiTree *lchild; struct BiTree *rchild; }BiTree; void main() { BiTree *root; queue<BiTree *> q; BiTree *t; q.push(root); int size; //核心代码 while(!q.empty()) { size=q.size(); while(size--) { t=q.front(); q.pop(); printf("%d ",t->val); if(t->lchild!=NULL) q.push(t->lchild); if(t->rchild!=NULL) q.push(t->rchild); } } }
Binary Tree Level Order Traversal
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.
/** * 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<vector<int> > levelOrder(TreeNode *root) { vector<int>level; vector<vector<int>>res; int size; TreeNode *t; if(root==NULL) return res; queue<TreeNode*> q; q.push(root); while(!q.empty()) { size=q.size(); while(size--) { t=q.front(); q.pop(); level.push_back(t->val); if(t->left!=NULL) q.push(t->left); if(t->right!=NULL) q.push(t->right); } res.push_back(level); level.clear(); } return res; } };
leetcode-Binary Tree Level Order Traversal 二叉树层序遍历
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