Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree{1,#,2,3},
1 2 / 3
return
[1,3,2].Note: Recursive solution is trivial, could you do it iteratively?
二叉树的非递归中序遍历。public List<Integer> inorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode node = root; while(node != null || !stack.isEmpty()) { while(node != null) { stack.push(node); node = node.left; } TreeNode curNode = stack.pop(); list.add(curNode.val); node = curNode.right; } return list; }
原文：http://www.cnblogs.com/cynchanpin/p/6744970.html