leetcode(NOWCODER)---binary-tree-postorder-traversal

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本题知识点： 树 leetcode
算法知识视频讲解
题目描述

Given a binary tree, return the postorder traversal of its nodes’ values.

For example:
Given binary tree{1,#,2,3},

1

2
/
3

return[3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

方案一（简单递归）：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    private ArrayList<Integer> list = new ArrayList<Integer>();
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        postOrder(root);
        return list;
    }
    private void postOrder(TreeNode root){
        if(root != null){
            postOrder(root.left);
            postOrder(root.right);
            list.add(root.val);
        }
    }
}

方案二（借助栈）：

按照根、左孩子、右孩子的顺序入栈，弹出根，插入到list首部，再弹出有孩子和左孩子，插入到list的首部，这样list中的顺序恰好就是后序遍历的顺序了

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    private ArrayList<Integer> list = new ArrayList<Integer>();
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        if(root == null){
            return list;
        }
        ArrayDeque<TreeNode> stack = new ArrayDeque<TreeNode>();
        stack.push(root);
        while(stack.size() > 0){
            TreeNode node = stack.pop();
            list.add(0, node.val);
            if(node.left != null){
                stack.push(node.left);
            }
            if(node.right != null){
                stack.push(node.right);
            }
        }
        return list;
    }
}