LeetCode 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]

.

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

分析：中序遍历的顺序是左右根，用堆栈迭代的方法模拟递归实现，堆栈用LinkedList实现

如果当前结点不为null，则压入堆栈，然后继续观察其左孩子；如果没有左孩子，则弹出堆栈中第一个数据，观察其右孩子。

代码：

/**
* 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) {
List<Integer> listVal = new ArrayList<Integer>();
LinkedList<TreeNode> listNode = new LinkedList<TreeNode>();
if(root == null) return listVal;
TreeNode p = root;

while(!listNode.isEmpty() || p != null){

if(p != null){
listNode.push(p);
p = p.left;
continue;
}
TreeNode t = listNode.pop();//
listVal.add(t.val);
p = t.right;

}

return listVal;
}
}