94. Binary Tree Inorder Traversal

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].

思路:和上题一样，还是不用递归，用stack实现

代码如下(已通过leetcode)

public static void inorder(List<Integer> list,TreeNode root) {
if(root==null) return;
Stack<TreeNode> stack=new Stack<TreeNode>();
while(root!=null||!stack.isEmpty()) {
while(root!=null) {
stack.push(root);
root=root.left;
}
root=stack.pop();
list.add(root.val);
root=root.right;
}
}