BM26 求二叉树的层序遍历

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