LintCode Binary Tree Inorder Traversal 二叉树的中序遍历（非递归）

给出一棵二叉树,返回其中序遍历。

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

样例

给出二叉树 {1,#,2,3},

1

\

2

/

3

返回 [1,3,2].

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Inorder in ArrayList which contains node values.
*/
public ArrayList<Integer> inorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode p = root;
while(p != null || !stack.empty()) {
while(p != null) {
stack.push(p);
p = p.left;
}
if(!stack.empty()) {
p = stack.pop();
list.add(p.val);
p = p.right;
}
}
return list;
}
}