Leetcode 606. Construct String from Binary Tree

Leetcode 606. Construct String from Binary Tree ———————————————–

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.

思路:
本题的考点是先序遍历,在遍历的过程中,注意什么时候加括号,什么时候不加括号。规律如下:

左子树空,右子树空: 啥都不加
左不空,右不空: 左加括号,右加括号
左空,右不空: 左加括号,右加括号
左不空,右空:左加括号

总结一下,只要右不空,就都加
如果右空,则左不空就左加

Java代码如下:


public class ConstructStringFromTree {

    public String tree2str(TreeNode t) {
        if(t == null) {
            return "";
        }

        StringBuilder sb = new StringBuilder();
        sb.append(t.val);

        if(t.right != null) {
            sb.append("(").append(tree2str(t.left)).append(")")
            .append("(").append(tree2str(t.right)).append(")");
        }else if(t.left != null) {
            sb.append("(").append(tree2str(t.left)).append(")");
        }

        return sb.toString();
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        TreeNode node1 = new TreeNode(1);
        TreeNode node2 = new TreeNode(2);
        TreeNode node3 = new TreeNode(3);
        TreeNode node4 = new TreeNode(4);

        node1.left = node2;
        node1.right = node3;
        node2.right = node4;

        ConstructStringFromTree c = new ConstructStringFromTree();
        String s = c.tree2str(node1);
        System.out.println(s);
    }

}

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