LeetCode 145. Binary Tree Postorder Traversal

原题链接在这里：https://leetcode.com/problems/binary-tree-postorder-traversal/

题目：

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?

题解：

类似Binary Tree Inorder Traversal 和 Binary Tree Preorder Traversal.

Method 1:  Recursion

Time Complexity: O(n). Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    //Method 1: Recursion
    public List<Integer> postorderTraversal(TreeNode root) {
        //Recursion
        List<Integer> ls = new ArrayList<Integer>();
        helper(root,ls);
        
        return ls;
    }
    private void helper(TreeNode root,List<Integer> ls){
        if(root == null){
            return;
        }
        helper(root.left,ls);
        helper(root.right,ls);
        ls.add(root.val);
    }
}

Method 2: 

Pretty much like method 2 of Binary Tree Preorder Traversal.

The difference is that when pushing into stack, push from left to right.

Time Complexity: O(n).

Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        Stack<TreeNode> stk = new Stack<>();
        stk.push(root);
        while(!stk.isEmpty()){
            TreeNode cur = stk.pop();
            res.add(cur.val);
            if(cur.left != null){
                stk.push(cur.left);
            }
            
            if(cur.right != null){
                stk.push(cur.right);
            }
        }
        
        Collections.reverse(res);
        return res;
    }
}

Method 3: Iteration + Stack

和Preorder, Inorder 类似.

当栈顶点右侧为空 或者 右侧已经traverse 过了, 就可以把栈顶的点加入res中. 节点pre用来记载刚刚加入res的点, 可以用来判定右侧是否已经traverse过了.

top.right == pre 说明right已经加入res里了.

Time Complexity: O(n), 每个点访问了一次. Space:O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stk = new Stack<TreeNode>();
        TreeNode pre = null;
        while(root != null || !stk.isEmpty()){
            if(root != null){
                stk.push(root);
                root = root.left;
            }else{
                TreeNode top = stk.peek();
                if(top.right == null || top.right == pre){
                    res.add(top.val);
                    stk.pop();
                    pre = top;
                }else{
                    root = top.right;
                }
            }
        }
        return res;
    }
}