二叉树的序列化
为了方便构造二叉树来验证我们的算法,这里先介绍下二叉树的序列化和反序列化。
序列化
先序遍历整颗二叉树即可,碰到节点存储下来,逗号隔开,若是空节点则记为#
。
代码如下:
public static String serialize(TreeNode root) {
StringBuilder sb = new StringBuilder();
sb.append('[');
serialize(root, sb);
sb.deleteCharAt(sb.length() - 1);
sb.append(']');
return sb.toString();
}
private static void serialize(TreeNode root, StringBuilder sb) {
if (root == null) {
sb.append('#');
sb.append(',');
} else {
sb.append(root.val);
sb.append(',');
serialize(root.left, sb);
serialize(root.right, sb);
}
}
反序列化
和序列化思路一致,先构造根节点,依次构造树的左子树、右子树。
因为int是栈空间的变量,而Integer是不可变类,我们这里构造一个包装类IntegerHolder
传递不停自增的数组下标。
代码如下:
public static TreeNode deserialize(String str) {
str = str.substring(1, str.length() - 1);
if (str.isEmpty()) return null;
String[] nums = str.split(",");
return deserialize(nums, new IntegerHolder());
}
private static TreeNode deserialize(String[] nums, IntegerHolder integerHolder) {
if (nums.length <= integerHolder.index) return null;
if (nums[integerHolder.index].equals("#")) {
integerHolder.index++;
return null;
}
TreeNode root = new TreeNode(Integer.parseInt(nums[integerHolder.index++]));
root.left = deserialize(nums, integerHolder);
root.right = deserialize(nums, integerHolder);
return root;
}
private static class IntegerHolder {
int index = 0;
}
二叉树的三种遍历
先序遍历(PreOrder)
递归方法
public void preOrder(TreeNode root) {
if (root == null) return;
System.out.print(root.val + " ");
preOrder(root.left);
preOrder(root.right);
}
迭代方法
public void preOrder(TreeNode root) {
if (root == null) return;
Stack<TreeNode> st = new Stack<>();
TreeNode node = root;
while (!st.empty() || node != null) {
if (node == null) {
node = st.pop();
} else {
System.out.print(node.val + " ");
if (node.right != null) st.push(node.right);
node = node.left;
}
}
}
中序遍历(InOrder)
递归方法
public void inOrder(TreeNode root) {
if (root == null) return;
inOrder(root.left);
System.out.print(root.val + " ");
inOrder(root.right);
}
迭代方法
public void inOrder(TreeNode root) {
if (root == null) return;
Stack<TreeNode> st = new Stack<>();
TreeNode node = root;
while (!st.empty() || node != null) {
if (node == null) {
node = st.pop();
System.out.print(node.val + " ");
node = node.right;
} else {
st.push(node);
node = node.left;
}
}
}
后序遍历(PostOrder)
递归方法
public void postOrder(TreeNode root) {
if (root == null) return;
postOrder(root.left);
postOrder(root.right);
System.out.print(root.val + " ");
}
迭代方法
public void postOrder(TreeNode root) {
if (root == null) return;
Stack<TreeNode> st = new Stack<>();
List<Integer> list = new ArrayList<>();
TreeNode node = root;
while (!st.empty() || node != null) {
if (node == null) {
node = st.pop();
} else {
list.add(node.val);
if (node.left != null) st.push(node.left);
node = node.right;
}
}
Collections.reverse(list);
for (int i : list) System.out.print(i + " ");
}