LeetCode: Binary Tree Inorder Traversal 解题报告

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.

SOL:

包括递归与非递归方法：

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<Integer> inorderTraversal1(TreeNode root) {
         List<Integer> ret = new ArrayList<Integer>();
         rec(root, ret);
         return ret;
     }

     public void rec(TreeNode root, List<Integer> ret) {
         if (root == null) {
             return;
         }

         rec(root.left, ret);
         ret.add(root.val);
         rec(root.right, ret);
     }

     public List<Integer> inorderTraversal(TreeNode root) {
         List<Integer> ret = new ArrayList<Integer>();
         if (root == null) {
             return ret;
         }

         Stack<TreeNode> s = new Stack<TreeNode>();
         TreeNode cur = root;

         while (true) {
             while (cur != null) {
                 s.push(cur);
                 cur = cur.left;
             }

             if (s.isEmpty()) {
                 break;
             }

             cur = s.pop();
             ret.add(cur.val);

             cur = cur.right;
         }

         return ret;
     }
 }