94.Binary Tree Inorder Traversal

Posted on | In

Question

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

Example:

1
2
3
4
5
6
7
8
Input: [1,null,2,3]
   1
    \
     2
    /
   3
Output: [1,3,2]

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

Answer

Recursive
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; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        return recursiveInTraversal(root, res);
    }
    public List<Integer> recursiveInTraversal(TreeNode root, List<Integer> res){
        if(root!=null){
            recursiveInTraversal(root.left, res);
            res.add(root.val);
            recursiveInTraversal(root.right, res);
        }
        return res;
    }
}

Running time: O(n), 1ms

Iterative
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
/**
 * 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> res = new ArrayList<>();
        if(root == null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while(node!=null||!stack.isEmpty()){
            while(node!=null){
                stack.push(node);
                node = node.left;
            }
            if(!stack.isEmpty()){
                node = stack.pop();
                res.add(node.val);
                node = node.right;
            }
        }
        return res;
    }
}

Running time: O(n), 2ms