[leetcode]Binary Tree InorderTraversal

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

算法思路：

1. 递归实现

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     List<Integer> res = new ArrayList<Integer>();
12     public List<Integer> inorderTraversal(TreeNode root) {
13         if(root == null) return res;
14         if(root.left != null) inorderTraversal(root.left);
15         res.add(root.val);
16         if(root.right != null) inorderTraversal(root.right);
17         return res;
18     }
19 }

思路2：

非递归版本：

借用栈，将当前节点的所有左节点（左节点的左节点....）压栈，然后依次弹栈处理当前节点并处理右子树

代码如下：

 1 public class Solution {
 2     public List<Integer> inorderTraversal(TreeNode root) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         if(root == null) return res;
 5         Stack<TreeNode> stack = new Stack<TreeNode>();
 6         TreeNode current = root;
 7         while(true){
 8             while(current != null){
 9                 stack.push(current);
10                 current = current.left;
11             }
12             if(stack.isEmpty()) break;
13             current = stack.pop();
14             res.add(current.val);
15             current = current.right;
16         }
17         return res;
18     }
19 }