树（二）二叉树递归和非递归遍历

1、递归遍历

// 先序遍历（递归实现）-------------------------------------------------------------   /*  1. Visit the node.  1. Call itself to traverse the node’s left subtree.      3. Call itself to traverse the node’s right subtree.  4. base case: there is no node  */   private void preOrder(Node localRoot){      if(localRoot != null)         {         visit(localRoot);       // System.out.print(localRoot.iData + " ");         preOrder(localRoot.leftChild);         preOrder(localRoot.rightChild);         }      }// 中序遍历（递归实现）-------------------------------------------------------------   private void inOrder(Node localRoot){      if(localRoot != null)         {         inOrder(localRoot.leftChild);         visit(localRoot);       //System.out.print(localRoot.iData + " ");         inOrder(localRoot.rightChild);         }      }// 后序遍历（递归实现）-------------------------------------------------------------   private void postOrder(Node localRoot){      if(localRoot != null)         {         postOrder(localRoot.leftChild);         postOrder(localRoot.rightChild);         visit(localRoot);       //System.out.print(localRoot.iData + " ");         }      }// -------------------------------------------------------------

2、非递归遍历

总体思想是：因为递归的本质是用栈实现的，所以写出非递归的形式要用到栈。入栈的顺序与访问的顺序相反，如中序遍历访问顺序是left,current,right,所以入栈的顺序相反是right,current,left
steo0:初始化当前节点current
step1:初始化栈
step3:出栈,并访问
Note：中序遍历和后序遍历需要先判断bPushed为true时访问
step2:入栈
循环结束条件是：栈为空

//算法框架 pushNode是入栈的方式public void templet(Node root){  //special case  if(root==null)   return;     //inital current node  Node current=root;    //initial stack  Stack s=new Stack();  pushNode(current,s);    //loop: base case is stack is empty  while( !s.empty() ){     //pop current=s.pop();  //visit or push  }//end while  }//end templet()

前序遍历（即深度优先搜索）

//current left right //step1:push in stack push current's right childpush current's left childvisit current node//step2:pop off stack//step3: loop step1-step2private void pushPreOrderNode(Node current, Stack s){   Node leftChild=current.leftChild;   Node rightChild=curret.rightChild;        if(rightChild!=null)    //push rightChild      s.push(rightChild);   if(leftChild!=null)     //push leftChild      s.push(leftChild);   visit(current);        //current always first visit in preOrder  }//end pushPreOrderNode()public void preOrder(Node root){    //special caseif(root==null)  return ;//initial current nodeNode current=root;//inintial stackStack s=new Stack();pushPreOrderNode(current,s);//base case:stack is emptywhile( !s.empty() ){  //pop  current=s.pop();  //push  pushPreOrderNode(current,s);}//end while}//end preOrder()

中序遍历

left,current,right

public void inOrder( Node root){   //special case   if(root==null)       return;   Node current=root;//initial stackStack s=new Stack();pushInOrderNode(current,s);//pop and visitwhile( !s.empty() ){ current=s.pop(); if(current.bPushed)visit(current); elsepushInOrderNode(current,s);}//end while}//end postOrder()private void pushInOrderNode(Node current,Stack s){   Node leftChild=current.leftChild;   Node rightChild=curret.rightChild;      if(rightChild!=null){   //push rightChild  s.push(rightChild);  rightChild.bPushed=false;   }//end if      s.push(current);        //push current      if(leftChild!=null){    //push leftChild  s.push(leftChild);  leftChild.bPushed=false;   }//end if     //set flag, ture present current's left and right has both pushed   current.bPushed=true;   }//end if}//end pushInOrderNode()

后序遍历

left right current
step1:初始化栈
按照step2初始化栈
step2:入栈
入栈顺序：current,right,left
如果current的中左右节点都入栈了，将current标识为true；
将入栈的左右节点标识初始化为false
step3:出栈
出栈节点设置为current,
如果current标识为ture,访问之；如果为false，做step2步骤
step4:重复step2&step3

public void postOrder( Node root){   //special case   if(root==null)       return;   Node current=root;//initial stackStack s=new Stack();pushPostOrderNode(current,s);//pop and visitwhile( !s.empty() ){     current=s.pop(); if(current.bPushed)    visit(current); else    pushPostOrderNode(current,s);}//end while}//end postOrder()// 左右子树尚未入栈，则依次将 根节点,右节点,左节点 入栈private void  pushPostOrderNode(Node current, Stack s){   Node leftChild=current.leftChild;   Node rightChild=curret.rightChild;      s.push(current);        //push current   if(rightChild!=null){   //push rightChild  s.push(rightChild);  rightChild.bPushed=false;   }//end if   if(leftChild!=null){    //push leftChild  s.push(leftChild);  rightChild.bPushed=false;   }//end if     //set flag, ture present current's left and right has both pushed   current.bPushed=true;}//end pushPostOrderNode()

3 深度优先搜索和层次遍历（广度优先搜索）

先序遍历即深度优先搜索，层次遍历即广度优先搜索

// -------------------------------------------------------------/*二叉树的深度优先搜索即先序遍历，用栈实现（非递归实现）*/// -------------------------------------------------------------/*层次搜索，即广度优先搜索step1:specail caseroot==null;step2:initial currentcurrent=root;step3:initial queuestep4:remove node in queue, and visit the current nodestep5:add node in queuestep6:loop step4 and step5base case: queue is empty*/public void gfs(Noode root){  //step1:specail case  if(root==null)    return ;    //step2:initial current  Node current=root;    //step3:initial queue  Queue q=new Queue();  q.add(current);    //step6:loop step4 and step5  //base case: queue is empty  while( !q.empty() ){    //step4:remove node in queuecurrent=q.remove();visit(current);Node leftChild=current.leftChild;Node rightChild=curret.rightChild;//step5:add node in queueif(leftChild!=null)   q.add(leftChild);if(rightChild!=null)   q.add(rightChild);  }//end while }//gfs()