【树】94. 二叉树的中序遍历

给定一个二叉树的根节点 root ，返回它的 中序 遍历。

 

示例 1：

输入：root = [1,null,2,3]
输出：[1,3,2]

方法一：
递归

/**
 * 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> rev = new ArrayList<Integer>();

        inorder(root,rev);
        return rev;
    }
    public void inorder(TreeNode root,List<Integer> list)
    {
        if(root!=null)
        {
            inorder(root.left,list);
            list.add(root.val);
            inorder(root.right,list);
        }
    }

}

 

方法二
非递归：空间复杂度O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

    //非递归中序遍历：栈

    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        //stack.push(root);
        while(root!=null || !stack.isEmpty()){
            //先遍历到最左下节点
            while(root!=null){
                stack.push(root);
                root = root.left;
            }

            //遍历根节点
            root = stack.pop();
            res.add(root.val);
            //遍历右子树
            root = root.right;
        }
        
        return res;
    }
}

 

方法三
非递归：线索二叉树。空间复杂度O(1)

class Solution {


    //中序线索二叉树

    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        TreeNode predecessor = null;

        while (root != null) {
            if (root.left != null) {
                // predecessor 节点就是当前 root 节点向左走一步，然后一直向右走至无法走为止
                predecessor = root.left;
                while (predecessor.right != null && predecessor.right != root) {
                    predecessor = predecessor.right;
                }
                
                // 让 predecessor 的右指针指向 root，继续遍历左子树
                if (predecessor.right == null) {
                    predecessor.right = root;
                    root = root.left;
                }
                // 说明左子树已经访问完了，我们需要断开链接
                else {
                    res.add(root.val);
                    predecessor.right = null;
                    root = root.right;
                }
            }
            // 如果没有左孩子，则直接访问右孩子
            else {
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}