思路：
逐层访问，通过访问当前层来得到下一层的节点。结束标志是下一层没有节点。
Go
package main
import . "nc_tools"
/*
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
/**
*
* @param root TreeNode类
* @return int整型二维数组
*/
func levelOrder( root *TreeNode ) [][]int {
// write code here
result := make([][]int,0)
if root == nil{
return result
}
curSet := make([]*TreeNode,0)
curSet = append(curSet,root)
tmpResult := make([]int,0)
for _,node := range curSet{
tmpResult = append(tmpResult,node.Val)
}
result = append(result,tmpResult)
for len(curSet) > 0 {
next := make([]*TreeNode,0)
tmpResult = make([]int,0)
for _,node := range curSet{
if node.Left != nil{
next = append(next,node.Left)
tmpResult = append(tmpResult,node.Left.Val)
}
if node.Right != nil{
next = append(next,node.Right)
tmpResult = append(tmpResult,node.Right.Val)
}
}
if len(tmpResult) > 0{
result = append(result,tmpResult)
}
curSet = next
}
return result
}
运行时间 8ms
占用内存 1168KB
C++
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
#include <iterator>
class Solution {
public:
/**
*
* @param root TreeNode类
* @return int整型vector<vector<>>
*/
vector<vector<int> > levelOrder(TreeNode* root) {
// write code here
vector<vector<int>> res;
if(root == nullptr){
return res;
}
vector<TreeNode*> curLevel;
curLevel.push_back(root);
vector<int> tmpResult;
tmpResult.push_back(root->val);
res.push_back(tmpResult);
while(curLevel.size() > 0){
tmpResult = vector<int>();
vector<TreeNode*> nextLevel;
for(auto &ele: curLevel){
if(ele->left != nullptr){
tmpResult.push_back(ele->left->val);
nextLevel.push_back(ele->left);
}
if(ele->right != nullptr){
tmpResult.push_back(ele->right->val);
nextLevel.push_back(ele->right);
}
}
if(tmpResult.size() > 0){
res.push_back(tmpResult);
}
curLevel = nextLevel;
}
return res;
}
};
运行时间 5ms
占用内存 516KB
https://clang.llvm.org/extra/clang-tidy/checks/modernize/loop-convert.html
原文链接: https://www.cnblogs.com/starter-songudi/p/17089406.html
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