94. Binary Tree Inorder Traversal java solutions

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

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

return [1,3,2].

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

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> ans = new ArrayList<Integer>();
        Stack<TreeNode> s = new Stack<TreeNode>();

        if(root == null) return ans;
        s.push(root);
        while(!s.isEmpty()){
            TreeNode node = s.peek();
            if(node.left != null){
                s.push(node.left);
                node.left = null;//原先这里没有设置为null一直内存超出限制，过不去。不知道为什么，
            }else{
                ans.add(node.val);
                s.pop();
                if(node.right != null){
                    s.push(node.right);
                }
            }
        }
        
        return ans;
    }
}