演算法題(三十七):按之字形順序列印二叉樹

純潔的程式碼發表於2020-04-03

題目描述
請實現一個函式按照之字形列印二叉樹,即第一行按照從左到右的順序列印,第二層按照從右至左的順序列印,第三行按照從左到右的順序列印,其他行以此類推。

分析
用兩個棧來實現,先把根結點放入s1,當行數為偶數時,s2從左到右放結點;當行數為奇數時,s1從右到左放結點;

筆者用java寫的程式不能100%AC,但c++卻可以,有大神可以指出來哪裡有問題嗎?

程式碼一
import java.util.ArrayList;
import java.util.Stack;

public class ZhiZiPrint {

public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode node1 = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node3 = new TreeNode(3);
TreeNode node4 = new TreeNode(4);
TreeNode node5 = new TreeNode(5);
TreeNode node6 = new TreeNode(6);
TreeNode node7 = new TreeNode(7);
TreeNode node8 = new TreeNode(8);
node1.left = node3;
node1.right = node2;
node3.left = node4;
node3.right = node5;
node2.left = node6;
node2.right = node7;
node7.left = node8;
ZhiZiPrint test = new ZhiZiPrint();
ArrayList<ArrayList<Integer> > lists = test.Print(node1);
for(ArrayList<Integer> list: lists){
for(int num: list){
System.out.println(num);
}
}
}

public ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) {
if(pRoot == null){
return null;
}
ArrayList<ArrayList<Integer> > lists = fun(pRoot);
return lists;
}

public ArrayList<ArrayList<Integer>> fun(TreeNode root){
ArrayList<ArrayList<Integer> > lists = new ArrayList<>();
Stack<TreeNode> s1 = new Stack<>();
Stack<TreeNode> s2 = new Stack<>();
s1.push(root);
while(!s1.isEmpty() || !s2.isEmpty()){
if(!s1.isEmpty()){
ArrayList<Integer> list = new ArrayList<>();
int len1 = s1.size();
for(int i=0; i<len1; i++){
TreeNode node = s1.peek();
list.add(node.val);
s1.pop();
if(node.left != null){
s2.push(node.left);
}
if(node.right != null){
s2.push(node.right);
}
}

lists.add(list);
}
if(!s2.isEmpty()){
ArrayList<Integer> list = new ArrayList<>();
int len2 = s2.size();
for(int i=0; i<len2; i++){
TreeNode node = s2.pop();
list.add(node.val);
if(node.right != null){
s1.push(node.right);
}
if(node.left != null){
s1.push(node.left);
}
}

lists.add(list);
}

}

return lists;
}

}
程式碼二(來源於牛客網)
連結:www.nowcoder.com/questio ...…
來源:牛客網

//思路2:利用兩個棧分別儲存奇數行和偶數行
class Solution {
public:
vector<vector<int> > Print(TreeNode* pRoot) {
vector<vector<int>> res;
if(!pRoot)return res;
stack<TreeNode*> sta1; //建立兩個棧,棧1用於存放奇數行節點,棧2用於存放偶數行節點
stack<TreeNode*> sta2;
sta1.push(pRoot);
vector<int> vec; //行容器,用於存入當前行輸出的結果
while(!sta1.empty()||!sta2.empty()){
if(sta2.empty()&&!sta1.empty()){
int len_1=sta1.size();
for(int i=0;i<len_1;i++){
TreeNode* tmp=sta1.top();
vec.push_back(tmp->val);
sta1.pop();
if(tmp->left)sta2.push(tmp->left); //棧2存放偶數行節點,按照從左子節點到右子節點的順序push
if(tmp->right)sta2.push(tmp->right);
}
res.push_back(vec);
vec.clear();
}

if(sta1.empty()&&!sta2.empty()){
int len_2=sta2.size();
for(int i=0;i<len_2;i++){
TreeNode* tmp=sta2.top();
vec.push_back(tmp->val);
sta2.pop();
if(tmp->right)sta1.push(tmp->right); //棧1存放奇數行節點,按照從右子節點到左子節點的順序push
if(tmp->left)sta1.push(tmp->left);
}
res.push_back(vec);
vec.clear();
}
}
return res;
}
};

---------------------
作者:另一個我竟然存在
來源:CSDN
原文:blog.csdn.net/qq_24034545…
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