Binary Tree Inorder Traversal

发表于   |   分类于 ，

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

Note:
Recursive solution is trivial, could you do it iteratively?

描述

给出一棵二叉树，返回中序遍历的结果

分析

非递归，使用 Stack 来模拟递归的方式

代码

递归：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> arrayList = new ArrayList<Integer>();

        if (root != null) {
            arrayList.addAll(inorderTraversal(root.left));
            arrayList.add(root.val);
            arrayList.addAll(inorderTraversal(root.right));
        }

        return arrayList;
    }
}

非递归：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> arrayList = new ArrayList<Integer>();

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;

        while (!stack.empty() || p != null) {

            if (p != null) {
                stack.push(p);
                p = p.left;
            } else {
                TreeNode pn = stack.pop();
                arrayList.add(pn.val);
                p = pn.right;
            }
        }

        return arrayList;
    }
}