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?

// Iterative 先用一个stack把中间节点存下来，左边如果为空就pop出来存进result然后换到right child

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


    }
}

// Recursive
public class Solution {
    public List inorderTraversal(TreeNode root) {
        ArrayList result = new ArrayList<>();
        inorder(root, result);
        return result;
    }
    public void inorder(TreeNode root, ArrayList result){
        if(root == null){
            return;
        }
        inorder(root.left, result);
        result.add(root.val);
        inorder(root.right, result);
    }
}