[Leetcode] 94. Binary Tree Inorder Traversal

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

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

   1    \     2    /   3

return [1,3,2].

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Recursive

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<Integer> inorderTraversal(TreeNode root) {        ArrayList<Integer> result = new ArrayList<Integer>();        traverse(root, result);        return result;    }    private void traverse(TreeNode root, ArrayList<Integer> result){        if(root == null) return;        traverse(root.left, result);        result.add(root.val);        traverse(root.right, result);    }}

iterative

import java.util.ArrayList;import java.util.Stack;public class Solution {    public ArrayList<Integer> inorderTraversal(TreeNode root) {        ArrayList<Integer> result = new ArrayList<Integer>();        if(root == null) return result;        Stack<TreeNode> stack = new Stack<TreeNode>();        while(!stack.isEmpty() || root != null){            if(root != null){                stack.push(root);                root = root.left;            } else {                root = stack.pop();                result.add(root.val);                root = root.right;            }        }        return result;    }}

preorder

public List<Integer>  iterativePreorder(TreeNode root){    List<Integer> res = new LinkedList<Integer>();    if(root==null) return res;    Stack<TreeNode> stack = new Stack<TreeNode>();    Stack.push(root);    while(!stack.isEmpty()){        TreeNode top = stack.pop();        res.add(top.val);        if(root.right!=null) stack.push(top.right);        if(root.left!=null) stack.push(top.left);    }    return res;}

postorder

public List<Integer> iterativePostorder(TreeNode root){    List<Integer> res = new LinkedList<Integer>();    if(root==null) return res;    Stack<TreeNode> stack = new Stack<TreeNode>();    TreeNode lastNodeVisited = null;    while(!stack.isEmpty()||root!=null){        if(root!=null){            stack.push(root);            root = root.left;        }else{            TreeNode peekNode = stack.peek();            if(peeknode.right!=null && peeknode.right!=lastNodeVisited){                root = peeknode.right;            }else{                stack.pop();                res.add(peeknode.val);                lastNodeVisited = peeknode;           }        }    }    return res;}