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?

 

法I：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) {
        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>();
}

法II：循环。使用stack以及一个set标记当前节点是否访问过

/**
 * 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<TreeNode> s = new Stack<TreeNode>();
        TreeNode curNode = root;
        while(true){
            if(visitedSet.contains(curNode)){ //如果访问过，访问右儿子
                result.add(curNode.val);
                curNode = curNode.right;
            }

            while(curNode!=null){ //如果没有访问过，自己先进栈，然后是左儿子         
                s.push(curNode);
                visitedSet.add(curNode);
                curNode = curNode.left;
            }
            if(s.empty()) break;
            
            //出栈一个节点
            curNode = s.peek();
            s.pop();
        }
        
        return result;
    }
}