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

Example

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

中序遍历是根左右

1. recursion

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> ans = new ArrayList<Integer>();
        if (root == null) {
            return ans;
        }

        inorderTraversal(ans, root);
        return ans;
    }


    public void inorderTraversal(ArrayList<Integer> ans, TreeNode root) {
        if (root == null) {
            return;
        }

        inorderTraversal(ans, root.left);
        ans.add(root.val);
        inorderTraversal(ans, root.right);
    }
}

1. divide and conquer

分治方法一致，只是放结点的顺序改变

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> ans = new ArrayList<Integer>();

        if (root == null) {
            return ans;
        }

        ArrayList<Integer> left = inorderTraversal(root.left);
        ArrayList<Integer> right = inorderTraversal(root.right);

        ans.addAll(left);
        ans.add(root.val);
        ans.addAll(right);

        return ans;


    }


}

1. non-recursion

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> ans = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();

        TreeNode cur = root;

        while (cur != null || !stack.isEmpty()) {

            while(cur != null) {
                stack.add(cur);
                cur = cur.left;
            }

            cur = stack.pop();
            ans.add(cur.val);
            cur = cur.right;

        }

        return ans;
    }
}