94. Binary Tree Inorder Traversal (Java)

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

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

 

Act I: recursive

/**
 * 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) {
        if(root == null) return result;
        inorder(root);
        return result;
    }
    
    public void inorder(TreeNode root) {
        if(root.left!=null) inorder(root.left);
        result.add(root.val);
        if(root.right!=null) inorder(root.right); 
    }
    
    private List<Integer> result = new ArrayList<Integer>();
}

Method II: cycle. Use stack and a set mark whether the current node visited

/**
 * 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> result = new ArrayList<Integer>();
        Set<TreeNode> visitedSet = new HashSet<TreeNode>();//Set有HashSet和TreeSet两种实现
        if(root == null) return result;
        Stack<The TreeNode> = S new new Stack <the TreeNode> (); 
        the TreeNode curNode = the root;
         the while ( to true ) {
             IF (visitedSet.contains (curNode)) { // if accessed, access to the right son 
                result.add (curNode.val) ; 
                curNode = curNode.right; 
            } 

            the while (curNode =! null ) { // If you have not visited, advanced stack own, then left his son          
                s.push (curNode); 
                visitedSet.add (curNode); 
                curNode = curNode.left ; 
            } 
            IF(s.empty ()) BREAK ; 
            
            // pop a node 
            curNode = s.peek (); 
            s.pop (); 
        } 
        
        return Result; 
    } 
}