LeetCode-Construct String from Binary Tree

最新推荐文章于 2021-06-12 02:40:54 发布

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本文介绍了一种将二叉树通过前序遍历转换为包含节点值和括号表示层次关系的字符串的方法。重点在于递归地处理根节点、左子树和右子树，并省略不影响映射关系的空括号对。

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Description：
You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.

The null node needs to be represented by empty parenthesis pair “()”. And you need to omit all the empty parenthesis pairs that don’t affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:

Input: Binary tree: [1,2,3,4]
       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:

Input: Binary tree: [1,2,3,null,4]
       1
     /   \
    2     3
     \  
      4 

Output: "1(2()(4))(3)"

Explanation: Almost the same as the first example, 
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

题意：给定一颗二叉树，要求以前序遍历的方式返回遍历的结果，利用括号表示节点之间的层次关系；

解法：既然要求我们需要以前序遍历的方式，我们可以递归依次遍历根节点、左孩子节点和右孩子节点；假设结果字符串为S，算法描述如下；

递归时，将其节点值加入S
当左孩子节点存在时，S加入’(’，递归调用函数，函数返回时，S加入‘)’，即已完成遍历左子树
当左孩子节点不存在且右孩子节点存在时，S加入‘()’表示为空的左孩子节点；
当右孩子节点存在时，S加入’(’，递归调用函数，函数返回时，S加入‘)’，即已完成遍历右子树

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public String tree2str(TreeNode t) {
        if (t == null) {
            return "";
        }
        StringBuilder result = new StringBuilder();
        result.append(t.val);
        if (t.left != null) {
            result.append('(');
            result.append(tree2str(t.left));
            result.append(')');
        } else if (t.right != null) {
            result.append("()");
        }
        if (t.right != null) {
            result.append('(');
            result.append(tree2str(t.right));
            result.append(')');
        }
        return result.toString();
    }
}