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]
.
非递归方法写二叉树的中序遍历
/** * 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) { Stack<TreeNode> stack = new Stack<TreeNode>(); List<Integer> inorder = new ArrayList<Integer>(); if (root == null) { return inorder; } TreeNode curt = root; while (curt != null || !stack.empty()) { while (curt != null) { stack.push(curt); curt = curt.left; } curt = stack.pop(); inorder.add(curt.val); curt = curt.right; } return inorder; } }