递归

• dfs一遍
• 在指针第二次会到该节点时打印该节点

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

    private void dfs(TreeNode root){
        if(root == null)
            return;
        dfs(root.left);
        ans.add(root.val);
        dfs(root.right);
    }
}

迭代

• 用栈模拟递归的过程
• 先把最左子树全部放入栈中
• 同样的在第二次回到该节点时将其放入list中

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if(root == null)
            return ans;
        Stack<TreeNode> stk = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stk.isEmpty()){
            while(cur != null){
                stk.push(cur);
                cur = cur.left;
            }
            //第二次来到cur
            cur = stk.pop();
            ans.add(cur.val);
            cur = cur.right;
        }
        return ans;
    }
}