数据结构与算法(二叉树的遍历算法)

定义节点

class ListNode{
      ListNode left;
      ListNode right;
      int val;
      public ListNode(int value){
            this.val=value;
      }
}

实现深度优先遍历
这里其实有三种前序中序后序，而且可以使用递归和非递归的方法。
递归的话在深度遍历一般需要用栈。

public void dfs(ListNode node){
       if(node==null){
             System.out.print("empty tree");
             return;
       }
       Stack<ListNode> stack = new Stack<ListNode>();
       stack.push(node);
       //用循环的方式来，当stack不为空就一直进行下去
       while(!stack.isEmpty()){
             ListNode rnode = stack.pop();
             System.out.print(rnode.val);
             if(rnode.right!=null){
                   stack.push(rnode.right);
             }
             if(rnode.left!=null){
                   stack.push(rnode.left);
             }
       }
}

而对于广度优先搜索
需要用到双向队列deque

public void dfs(LsitNode node){
      if(node==null){
            System.out.print("empty tree"); 
            return;
      }
      ArrayDeque<ListNode> deque = new ArrayDeque<ListNode>();
      deque.add(node);
      while(!deque.isEmpty()){
      //remove() and add()
            ListNode rnode = deque.remove();
            System.out.print(rnode.val+"  ");
            if(rnode.left!=null){
                  deque.add(rnode.left);
            }
            if(rnode.right!=null){
                  deque.add(rnode.right);
            }
      }
}

问题：LeetCode的题，给一个二叉树的根节点root，返回它的中序遍历List
输入：root = [1,null,2,3]
输出：[1,3,2]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

解答：

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        // List用add,remove。而stack用pop,push
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(!stack.isEmpty() ||cur != null){
            if(cur != null){
                stack.push(cur);
                cur = cur.left;
            }else{
                cur = stack.pop();
                list.add(cur.val);
                cur = cur.right;
            }
        }
        return list;
    }
}