In a binary tree, a lonely node is a node that is the only child of its parent node. The root of the tree is not lonely because it does not have a parent node.
Given the root of a binary tree, return an array containing the values of all lonely nodes in the tree. Return the list in any order.
Example 1:
Input: root = [1,2,3,null,4]
Output: [4]
Explanation: Light blue node is the only lonely node.
Node 1 is the root and is not lonely.
Nodes 2 and 3 have the same parent and are not lonely.
Example 2:
Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2]
Output: [6,2]
Explanation: Light blue nodes are lonely nodes.
Please remember that order doesn't matter, [2,6] is also an acceptable answer.
Example 3:
Input: root = [11,99,88,77,null,null,66,55,null,null,44,33,null,null,22]
Output: [77,55,33,66,44,22]
Explanation: Nodes 99 and 88 share the same parent. Node 11 is the root.
All other nodes are lonely.
Example 4:
Input: root = [197]
Output: []
Example 5:
Input: root = [31,null,78,null,28] Output: [78,28]
怎么判断父母节点啊：利用了形态上的特殊性
有左节点的话，右节点就为空。所以其实也就是摆弄一下左右的关系。
参考：https://leetcode.com/problems/find-all-the-lonely-nodes/discuss/669635/Java-recursive-top-down-as-parent-passes-isLonely-to-each-children
public List<Integer> getLonelyNodes(TreeNode root) {
List<Integer> nodes = new ArrayList<>();
getLonelyNodes(root, false, nodes); // root is not lonely
return nodes;
}
private void getLonelyNodes(TreeNode root, boolean isLonely, List<Integer> nodes) {
if (root == null) return;
if (isLonely) {
nodes.add(root.val);
}
getLonelyNodes(root.left, root.right == null, nodes);
getLonelyNodes(root.right, root.left == null, nodes);
}