• 二叉树的四种遍历的非递归实现


    心血来潮,想实现一下二叉树的四种遍历非递归,就在

    发现自己的功夫真的下降,居然用了一个多小时才都在LeetCode上Accept

    基本功还需要再练习呀,时间长了不用都快忘了,不过还好理解了这个过程,回忆起来比较快。

    其实有个诀窍,就是脑海里模仿一遍动画过程,会记得比较快。

    不费话了,上代码:(具体思路和需要注意的地方见注释,以后决定都用英语写注释,剩的乱码,语法不通,请见谅)

    先序遍历(栈)

    
    public class Solution {
        public List<Integer> preorderTraversal(TreeNode root) {
            List<Integer> res = new ArrayList<Integer>();
            if(root == null) return res;
            LinkedList<TreeNode> s = new LinkedList<TreeNode>();
            TreeNode p = root;
            s.push(root);
            while(!s.isEmpty()){
                p = s.getFirst();
                s.pop();
                res.add(p.val);
                //because of FILO,must  push right node firstly,then push left node .
                if(p.right != null) {s.push(p.right);}
                if(p.left != null){s.push(p.left);}
            }
            return res;
        }
    }
    

    中序遍历(栈)

    
    public class Solution {
        public List<Integer> inorderTraversal(TreeNode root) {
            List<Integer> res = new ArrayList<Integer>();
            if(root == null) return res;
            LinkedList<TreeNode> s = new LinkedList<TreeNode>();
            TreeNode p = root;
            //Attention:the condition
            while(!s.isEmpty()||p!=null){
                if(p!=null){
                    s.push(p);
                    p = p.left;
                }else{
                    p = s.getFirst();
                    s.pop();
                    res.add(p);
                    p = p.right;
                }
            }
            return res;
        }
    }
    

    后序遍历(栈)

    相对麻烦一下,因为当弹出某个最左节点后,不能直接出其父亲节点,要先去遍历其父亲节点的右子树,所以使用双指针的思想记录之前访问的节点

    
    public class Solution {
        public List<Integer> postorderTraversal(TreeNode root) {
            List<Integer> res = new ArrayList<Integer>();
            if(root==null) return res;
            LinkedList<TreeNode> s = new LinkedList<TreeNode>();
            //double pointer,p is the current node , q is the pre node
            TreeNode p = root;
            TreeNode q = new TreeNode(-1);
            do{
                while(p!=null){//towards left-down
                    s.push(p);
                    p = p.left;
                }
                q = null;
                while(!s.isEmpty()){
                    p = s.getFirst();
                    s.pop();
                    //right child is visited or not exist
                    if(p.right == q){
                        res.add(p.val);
                        q = p;//save the node which is visted just now 
                    }else{
                        //current node does not need vist,push it again
                        s.push(p);
                        //deal the right child tree firstly
                        p = p.right;
                        break;
                    }
    
                }
            }while(!s.isEmpty());
            return res;
        }
    }
    

    层序遍历(队列)

    
    public class Solution {
        public List<List<Integer>> levelOrder(TreeNode root) {
            List<List<Integer>> res = new ArrayList<List<Integer>>();
            //we use LinkedList as a queue
            LinkedList<TreeNode> q = new LinkedList<TreeNode>();
            if(root == null) return res;
            TreeNode p = root;
            q.addLast(p);
            while(!q.isEmpty()){
                int tmpSize = q.size();
                List<Integer> tmpList = new ArrayList<Integer>();
                //Attention:i<tmpSize
                for(int i=0;i<tmpSize;i++){
                    p = q.getFirst();
                    q.poll();
                    tmpList.add(p.val);
                    if(p.left!=null)  q.addLast(p.left);
                    if(p.right!=null) q.addLast(p.right);
                }
                res.add(tmpList);
            }
            return res;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/yueyanglou/p/5274755.html
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