试题:
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?
代码:
递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
List<Integer> out = new ArrayList<>();
public List<Integer> postorderTraversal(TreeNode root) {
if(root==null) return out;
postorderTraversal(root.left);
postorderTraversal(root.right);
out.add(root.val);
return out;
}
}
迭代:使用root-right-left,再反转形式
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> out = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode tmp = stack.pop();
if(tmp==null) continue;
out.add(tmp.val);
stack.push(tmp.left);
stack.push(tmp.right);
}
Collections.reverse(out);
return out;
}
}