d*i
2 楼
请大侠给个java版本的完整代码, Google上找不到。。谢谢了。
我手上的代码按照leetcode写的。但是readNextToken不完整,而且isNumber永远都是
false,郁闷:
public static void serializeBinaryTree(TreeNode node, BufferedWriter out)
throws IOException {
if (node == null) {
out.write("# ");
} else {
out.write(node.val + " ");
serializeBinaryTree(node.left, out);
serializeBinaryTree(node.right, out);
}
}
public static void deserializeBinaryTree(TreeNode node,
BufferedReader in) {
int token;
boolean isNumber;
if (!readNextToken(token, in, isNumber)) {
return;
} else {
if (isNumber) {
node = new TreeNode(token);
deserializeBinaryTree(node.left, in);
deserializeBinaryTree(node.right, in);
}
}
}
我手上的代码按照leetcode写的。但是readNextToken不完整,而且isNumber永远都是
false,郁闷:
public static void serializeBinaryTree(TreeNode node, BufferedWriter out)
throws IOException {
if (node == null) {
out.write("# ");
} else {
out.write(node.val + " ");
serializeBinaryTree(node.left, out);
serializeBinaryTree(node.right, out);
}
}
public static void deserializeBinaryTree(TreeNode node,
BufferedReader in) {
int token;
boolean isNumber;
if (!readNextToken(token, in, isNumber)) {
return;
} else {
if (isNumber) {
node = new TreeNode(token);
deserializeBinaryTree(node.left, in);
deserializeBinaryTree(node.right, in);
}
}
}
d*e
3 楼
我是分开寄的
p*3
4 楼
觉得leetcode上解法很不好,要是binary value怎么办,node的值不是数值怎么办
y*n
6 楼
Any suggestion?
s*x
8 楼
at least instead of '#' for null, we should write extra byte with value of 1
or 0 to indicate the node is NULL or not.
or 0 to indicate the node is NULL or not.
d*i
10 楼
up
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