0
点赞
收藏
分享

微信扫一扫

力扣:二叉树的中序遍历

草原小黄河 2022-03-19 阅读 75
leetcode

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:

img

输入:root = [1,null,2,3]
输出:[1,3,2]

示例 2:

输入:root = []
输出:[]

示例 3:

输入:root = [1]
输出:[1]

示例 4:

img

输入:root = [1,2]
输出:[2,1]

示例 5:

img

输入:root = [1,null,2]
输出:[1,2]

提示:

  • 树中节点数目在范围 [0, 100]
  • -100 <= Node.val <= 100

进阶: 递归算法很简单,你可以通过迭代算法完成吗?

/**
 * 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 {
    List<Integer> list = new ArrayList<Integer>();
    public List<Integer> inorderTraversal(TreeNode root) {
        traversal(root);
        return list;    
    }
    public void traversal(TreeNode root){
        if(root==null)
            return;
        traversal(root.left);
        list.add(root.val);
        traversal(root.right);
    }
}
/**
 * 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 {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        Stack<TreeNode> s = new Stack<>();
        while(root!=null || !s.isEmpty()){
            while(root!=null){
                s.push(root);
                root = root.left;
            }
            root = s.pop();
            list.add(root.val);
            root = root.right;
        }
        return list;    
    }
}
举报

相关推荐

0 条评论