Binary Tree Inorder Traversal 二叉树中序遍历@LeetCode

中序遍历迭代方法在二叉树面试总结一文中写了

package Level2;

import java.util.ArrayList;
import Utility.TreeNode;

/**
 * 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}".
 *
 */
public class S24 {

	public static void main(String[] args) {

	}

	public ArrayList<Integer> inorderTraversal(TreeNode root) {
		ArrayList<Integer> ret = new ArrayList<Integer>();
		rec(root, ret);
		return ret;
    }
	
	public void rec(TreeNode root, ArrayList<Integer> ret){
		if(root == null){
			return;
		}
		
		rec(root.left, ret);
		ret.add(root.val);
		rec(root.right, ret);
	}

}

最优解法:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> ret = new ArrayList<Integer>();
		if(root == null){
			return ret;
		}
		
		Stack<TreeNode> stack = new Stack<TreeNode>();
		TreeNode cur = root;
		while(true){
			while(cur != null){
				stack.push(cur);
				cur = cur.left;
			}
			if(stack.isEmpty()){
				break;
			}
			cur = stack.pop();
			ret.add(cur.val);
			cur = cur.right;
		}
		return ret;
    }
}

之前自己写的，可以AC，但是要改变原树结构，不好！

public static ArrayList<Integer> inorderTraversal2(TreeNode root) {
		ArrayList<Integer> ret = new ArrayList<Integer>();
        if(root == null){
            return ret;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        TreeNode cur = stack.peek();
        while(!stack.isEmpty()){
            while(cur.left != null){
                stack.push(cur.left);
                cur = cur.left;
            }
            cur = stack.pop();
            ret.add(cur.val);
            if(cur.right != null){
                cur = cur.right;
                stack.push(cur);
            }else{
            	if(stack.isEmpty()){
            		break;
            	}
            	cur = stack.peek();
            	if(cur.left!=null){
            		cur.left = null;
            	}
            }
        }
        return ret;
    }