Total Accepted: 16494 Total Submissions: 47494
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.
【解題思路】
1、遞迴

Java AC

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

2、非遞迴 棧儲存節點，當左孩子不為空，一直入棧。否則出棧，同時將棧頂元素存入list中，針對當前節點右孩子做同樣操作。

Java AC

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