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题目地址：

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

For example:

1    \     2    /   3

return [1,3,2].

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

既然不让用递归，那就只能用堆栈了。

public class BinaryTreeInorderTraversal {
       public static List
   
     inorderTraversal(TreeNode root) {        List
    
      res = new ArrayList<>();        if (root == null)            return res;        Stack
     
       stack = new Stack<>();        TreeNode node = root;        while (node != null) {            stack.push(node);            node = node.left;        }        while (!stack.isEmpty()) {            node = stack.pop();            res.add(node.val);            if (node.right != null) {                node = node.right;                while (node != null) {                    stack.push(node);                    node = node.left;                }            }        }        return res;    }    public static void main(String[] args) {        TreeNode root = new TreeNode(1);        root.right = new TreeNode(2);        root.right.left = new TreeNode(3);        System.out.println(inorderTraversal(root));        return;    }}
     
    
   

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