标签:binary String Binary tree construct 二叉树 parenthesis stack string
[抄题]:
You need to construct a binary tree from a string consisting of parenthesis and integers.
The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root's value and a pair of parenthesis contains a child binary tree with the same structure.
You always start to construct the left child node of the parent first if it exists.
Example:
Input: "4(2(3)(1))(6(5))" Output: return the tree root node representing the following tree: 4 / \ 2 6 / \ / 3 1 5
Note:
- There will only be
'('
,')'
,'-'
and'0'
~'9'
in the input string. - An empty tree is represented by
""
instead of"()"
.
不用q的做法,就会要多引入变量,有点麻烦:https://leetcode.com/problems/construct-binary-tree-from-string/discuss/100355/Java-Recursive-Solution
用q的做法:https://leetcode.com/problems/construct-binary-tree-from-string/discuss/100359/Java-stack-solution
public class Solution { public TreeNode str2tree(String s) { Stack<TreeNode> stack = new Stack<>(); for(int i = 0, j = i; i < s.length(); i++, j = i){ char c = s.charAt(i); if(c == ')') stack.pop(); else if(c >= '0' && c <= '9' || c == '-'){ while(i + 1 < s.length() && s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9') i++; TreeNode currentNode = new TreeNode(Integer.valueOf(s.substring(j, i + 1))); if(!stack.isEmpty()){ TreeNode parent = stack.peek(); if(parent.left != null) parent.right = currentNode; else parent.left = currentNode; } stack.push(currentNode); } } return stack.isEmpty() ? null : stack.peek(); } }
标签:binary,String,Binary,tree,construct,二叉树,parenthesis,stack,string 来源: https://www.cnblogs.com/immiao0319/p/14942872.html
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