590. N-ary Tree Postorder Traversal
Problem
Given an n-ary tree, return the postorder traversal of its nodes' values.
For example, given a 3-ary tree:
Return its postorder traversal as: [5,6,3,2,4,1].
Note: Recursive solution is trivial, could you do it iteratively?
Solution (Recursion)
class Solution {
public List<Integer> postorder(Node root) {
List<Integer> res = new ArrayList<>();
helper(root, res);
return res;
}
private void helper(Node root, List<Integer> res) {
if (root == null) return;
if (root.children != null) {
for (Node child: root.children) {
helper(child, res);
}
}
res.add(root.val);
}
}
Solution (Iteration)
按顺序放入stack,正好level方面是从root到leaves,顺序方面是从最右到最左,因为stack是先入后出。这样最后reverse一下就是先左后右,先子后根。
巧妙。
class Solution {
public List<Integer> postorder(Node root) {
List<Integer> res = new ArrayList<>();
if (root == null) return res;
Stack<Node> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
Node node = stack.pop();
res.add(node.val);
for (Node child: node.children) {
stack.push(child);
}
}
Collections.reverse(res);
return res;
}
}
145. Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [3,2,1]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution (Iteration)
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if (root == null) return res;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
res.add(node.val);
if (node.left != null) stack.push(node.left);
if (node.right != null) stack.push(node.right);
}
Collections.reverse(res);
return res;
}
}
Solution (Recursion)
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
helper(root, res);
return res;
}
private void helper(TreeNode node, List<Integer> res) {
if (node == null) return;
if (node.left != null) helper(node.left, res);
if (node.right != null) helper(node.right, res);
res.add(node.val);
}
}
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