面试准备--二叉树的非递归遍历(Java版)

函数是写在树类中.

前序和中序都是差不多,不过在于输出的顺序而已.

1. 前序在遍历最左节点时,边遍历边输出.

2. 中序在取出栈顶元素和放入右子树之间时输出.

//中序非递归遍历二叉树
public ArrayList<Integer> zsort(){
Stack<Node> s = new Stack<Node>();
ArrayList<Integer> q = new ArrayList<Integer>();
Node p;
s.add(root);
while(!s.empty()){
p = s.peek();
while(p!=null){
s.push(p.getLeft());
p = p.getLeft();
}
s.pop();
if(!s.empty()){
p = s.pop();
q.add(p.getData());
s.push(p.getRight());
}
}
return q;
}

//非递归前序遍历二叉树
public ArrayList<Integer> qsort(){
Stack<Node> s = new Stack<Node>();
ArrayList<Integer> q = new ArrayList<Integer>();
Node p;
s.push(root);
while(!s.empty()){
p = s.peek();
while(p!=null){
q.add(p.getData());
s.push(p.getLeft());
p = p.getLeft();
}
s.pop();
if(!s.empty()){
p = s.pop();
s.push(p.getRight());
}
}
return q;
}

后序遍历比较麻烦,要保证左右子树必须在根节点之前遍历输出.遍历根节点的条件:无左右子树或者已经遍历了左右子树(遍历的前一个节点就是左右子树).右子树先进栈,这样在出栈的时候就是左子树先出栈.

//非递归后序遍历二叉树
public ArrayList<Integer> hsort(){
Stack<Node> s = new Stack<Node>();
ArrayList<Integer> q = new ArrayList<Integer>();
Node now = null;
Node pre = null;
s.push(root);
while(!s.empty()){
now = s.peek();
if((now.getLeft() == null && now.getRight() == null) ||
(pre != null &&( pre == now.getLeft() || pre == now.getRight() ) ) ){
q.add(now.getData());
s.pop();
pre = now;
}
else
{
if(now.getRight()!=null){
s.push(now.getRight());
}
if(now.getLeft()!=null){
s.push(now.getLeft());
}
}

}
return q;
}