【LeetCode】145. Binary Tree Postorder Traversal

Difficulty: Hard

Description

https://leetcode.com/problems/binary-tree-postorder-traversal/

Given a binary tree, return the postordertraversal of its nodes' values.

Example:

Input: [1,null,2,3]

   1

    \

     2

    /

   3

Output: [3,2,1]

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

Intuition

Method 1. Using one stack to store nodes, and another to store a flag wheather the node has traverse right subtree.

Method 2. Stack + Collections.reverse( list )

Solution

Method 1

    public List<Integer> postorderTraversal(TreeNode root) {

        List<Integer> list = new LinkedList<Integer>();

        Stack<TreeNode> nodeStk = new Stack<TreeNode>();

        Stack<Boolean> tag = new Stack<Boolean>();

        while(root!=null || !nodeStk.isEmpty()){

            while(root!=null){

                nodeStk.push(root);

                tag.push(false);

                root=root.left;

            }

            if(!tag.peek()){

                tag.pop();

                tag.push(true);

                root=nodeStk.peek().right;

            }else{

                list.add(nodeStk.pop().val);

                tag.pop();

            }

        }

        return list;

    }

Method 2

    public List<Integer> postorderTraversal(TreeNode root) {

        LinkedList<Integer> list = new LinkedList<Integer>();

        Stack<TreeNode> stk = new Stack<>();

        stk.push(root);

        while(!stk.isEmpty()){

            TreeNode node = stk.pop();

            if(node==null)

                continue;

            list.addFirst(node.val);  //LinkedList's method. If using ArrayList here,then using 'Collections.reverse(list)' in the end;

            stk.push(node.left);

            stk.push(node.right);

        }

        return list;

    }

Complexity

Time complexity : O(n)

Space complexity : O(nlogn)

What I've learned

1. linkedList.addFirst( e )

2. Collections.reverse( arraylist )

More:【目录】LeetCode Java实现

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