二叉树后续遍历的几种方式

标签：思路   lse   结果   arraylist   后序   数据结构   ||   problems   cti   

二叉树的后续遍历(Leetcode 145)

数据结构定义：

 // 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;
      }
  }

递归写法：

class Solution {
    List<Integer> result =new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null)
            return result;
        if(root.left != null)
            postorderTraversal(root.left);
        if(root.right != null)
            postorderTraversal(root.right);
        result.add(root.val);
        return result;
    }
}

普通迭代：

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result =new ArrayList<>();
        if(root == null)
            return result;
      //pre的作用是记忆上一次遍历的节点，以便判断是否有右子树或右子树未遍历
        TreeNode node = root,pre = null;
        Stack<TreeNode> stack =new Stack<>();
        while(!stack.isEmpty() || node != null){
            while(node != null){
                stack.push(node);
                node = node.left;
            }
            if((node=stack.peek()).right == null 
               || node.right == pre){
                pre = node;
                result.add(stack.pop().val);
                node =null;
            }else{
                pre = node;
                node =node.right;
            }
        }
        return result;
    }
}

借鉴前序遍历思想：

//思路：
//前序遍历的过程是：(根-左-右)
//那么我们遍历的过程变为：(根-右-左)，最后的结果再进行下倒置 
//顺序就变成了:(左-右-根),完成对后序的遍历
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result =new ArrayList<>();
        if(root == null){
            return result;
        }
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node =root;
        while(!stack.isEmpty() || node != null){
            while(node != null){
                stack.push(node);
                result.add(node.val);
                node = node.right;
            }
            node = stack.pop().left;
        }
        Collections.reverse(result);
        return result;
    }
}

莫里斯遍历：

//还是借鉴前序遍历的思路，莫里斯遍历可参看我的中序遍历一文
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if(root == null){
            return result;
        }
        TreeNode node =root,pre = null;
        while(node != null){
            if(node.right == null){
                result.add(node.val);
                node =node.left;
            }else{
                pre = node.right;
                while(pre.left != null && pre.left != node){
                    pre = pre.left;
                }
                if(pre.left == null){
                    pre.left = node;
                    result.add(node.val);
                    node = node.right;
                }else{
                    pre.left = null;
                    node =node.left;
                }
            }
        }
        Collections.reverse(result);
        return result;
    }

借鉴头指针插入法:

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        LinkedList<Integer> result =new LinkedList<>();
        LinkedList<TreeNode> stack =new LinkedList<>();
        if(root == null){
            return result;
        }
        stack.addFirst(root);
        while(!stack.isEmpty()){
            TreeNode node =stack.removeFirst();
            result.addFirst(node.val);
            if(node.left != null){
                stack.addFirst(node.left);
            }
            if(node.right != null){
                stack.addFirst(node.right);
            }
        }
        return result;
    }
}

二叉树后续遍历的几种方式

标签：思路   lse   结果   arraylist   后序   数据结构   ||   problems   cti   

原文地址：https://www.cnblogs.com/CodingXu-jie/p/13866761.html