94. Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],
   1
    \
     2
    /
   3
return [1,3,2].

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

##Approach 1: recursive

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

##Approach 2: iterative 采用stack来解。

1. 将root压入stack；
2. 若左孩子不为null，则不断将左孩子压入stack，直到为null；
3. 若stack不为空，则pop栈顶元素，为node；
4. 若右孩子不为null，则令node = node.right，压入stack；
5. 若左孩子不为空，则不断将左孩子压入stack，直到为null；
6. 重复step 3。

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

more concise version：

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