Binary Tree Inorder Traversal

Link: http://oj.leetcode.com/problems/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.


OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

Recusion Version:

 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     ArrayList<Integer> nodes = new ArrayList<Integer>();
12     public ArrayList<Integer> inorderTraversal(TreeNode root) {
13         inorderTraversal_recursion(root);
14         return nodes;
15     }
16     public void inorderTraversal_recursion(TreeNode root){
17         if(root==null)
18             return;
19         inorderTraversal_recursion(root.left);
20         nodes.add(root.val);
21         inorderTraversal_recursion(root.right);
22     }
23     
24 }

Iterative version:

 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     public ArrayList<Integer> inorderTraversal(TreeNode root) {
12         // the nodes is used to store the result
13         ArrayList<Integer> nodes = new ArrayList<Integer>();
14         if (root == null)
15             return nodes;
16         Stack<TreeNode> stack = new Stack<TreeNode>();
17         TreeNode node = root;
18         // note in the begin of the loop, stack is empty,
19         // so we need node != null to make sure the while-loop
20         // can be executed.
21         while (node != null || !stack.isEmpty()) {
22             // if the node is not null,
23             // we should get the left child
24             if (node != null) {
25                 stack.push(node);
26                 node = node.left;
27             }
28             // if the node is null
29             // i.e. reach the situation that the parent
30             // don't have left child.
31             // we should pop the top of the stack,
32             // and do the operation on the right child.
33             else {
34                 node = stack.pop();
35                 nodes.add(node.val);
36                 node = node.right;
37             }
38         }
39         return nodes;
40     }
41 }

 

posted on 2014-05-04 02:20  Atlas-Zzz  阅读(130)  评论(0)  编辑  收藏  举报

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