LeetCode - Easy - 606. Construct String from Binary Tree

朴灿烈づ我的快乐病毒、 2022-10-06 07:49 108阅读 0赞

## Topic ##

*  String
 *  Tree

## Description ##

[https://leetcode.com/problems/construct-string-from-binary-tree/][https_leetcode.com_problems_construct-string-from-binary-tree]

Given the `root` of a binary tree, construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way, and return it.

Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.

**Example 1:**

![b5429c455e49c30a8e2d64f7b61a076e.png][]

Input: root = [1,2,3,4]
    Output: "1(2(4))(3)"
    Explanation: Originallay it needs to be "1(2(4)())(3()())", but you need to omit all the unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)"

**Example 2:**

![c3ba168454879b9da02d4b4c86e99b40.png][]

Input: root = [1,2,3,null,4]
    Output: "1(2()(4))(3)"
    Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

**Constraints:**

*  The number of nodes in the tree is in the range  \[ 1 , 1 0 4 \] \[1, 10^4\] \[1,104\].
 *  `-1000 <= Node.val <= 1000`

## Analysis ##

方法一：我写的，递归法

方法二，别人写的，递归法

## Submission ##

import com.lun.util.BinaryTree.TreeNode;
    
    public class ConstructStringFromBinaryTree { 
    
    	//方法一：我写的，递归法
    	public String tree2str(TreeNode root) { 
    		return dfs(root).toString();
    	}
    
    	private StringBuilder dfs(TreeNode node) { 
    		StringBuilder sb = new StringBuilder();
    		if (node == null)
    			return sb;
    
    		sb.append(node.val);
    
    		if (node.left != null && node.right == null) { 
    			sb.append("(");
    			sb.append(dfs(node.left));
    			sb.append(")");
    		} else if (node.right != null) { 
    			sb.append("(");
    			sb.append(dfs(node.left));
    			sb.append(")(");
    			sb.append(dfs(node.right));
    			sb.append(")");
    		}
    		return sb;
    	}
    
    	//方法二，别人写的，递归法
    	public String tree2str2(TreeNode t) { 
    		StringBuilder sb = new StringBuilder();
    		preorder(t, sb);
    		return sb.toString();
    	}
    
    	private void preorder(TreeNode root, StringBuilder sb) { 
    		if (root == null) { 
    			return;
    		}
    		sb.append(root.val);
    		if (root.left != null) { 
    			sb.append("(");
    			preorder(root.left, sb);
    			sb.append(")");
    		}
    		if (root.right != null) { 
    			if (root.left == null) { 
    				sb.append("()");
    			}
    			sb.append("(");
    			preorder(root.right, sb);
    			sb.append(")");
    		}
    	}
    }

## Test ##

import static org.junit.Assert.*;
    import org.junit.Test;
    
    import com.lun.util.BinaryTree;
    
    public class ConstructStringFromBinaryTreeTest { 
    
    	@Test
    	public void test() { 
    		ConstructStringFromBinaryTree obj = new ConstructStringFromBinaryTree();
    
    		assertEquals("1(2(4))(3)", obj.tree2str(BinaryTree.integers2BinaryTree(1,2,3,4)));
    		assertEquals("1(2()(4))(3)", obj.tree2str(BinaryTree.integers2BinaryTree(1,2,3,null,4)));
    	}
    	
    	@Test
    	public void test2() { 
    		ConstructStringFromBinaryTree obj = new ConstructStringFromBinaryTree();
    		
    		assertEquals("1(2(4))(3)", obj.tree2str2(BinaryTree.integers2BinaryTree(1,2,3,4)));
    		assertEquals("1(2()(4))(3)", obj.tree2str2(BinaryTree.integers2BinaryTree(1,2,3,null,4)));
    	}
    }

[https_leetcode.com_problems_construct-string-from-binary-tree]: https://leetcode.com/problems/construct-string-from-binary-tree/
[b5429c455e49c30a8e2d64f7b61a076e.png]: /images/20221005/6e5e5a95cd254b659d9248056ede04ae.png
[c3ba168454879b9da02d4b4c86e99b40.png]: /images/20221005/0ad42b3984fa4898803a52198c6dfd23.png