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;
    }
}