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BinaryTreePostorderTraversal.java
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BinaryTreePostorderTraversal.java
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package algorithms;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
/**
* 145. Binary Tree Postorder Traversal
* https://leetcode.com/problems/binary-tree-postorder-traversal/
* Difficulty : Hard
* Related Topics : Stack, Tree, Depth-first Search
*
* 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?
*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*
* created by cenkc on 8/12/2020
*/
public class BinaryTreePostorderTraversal {
/**
* Runtime: 0 ms
* Memory Usage: 37.7 MB
*
* @param root
* @return
*/
// iterative
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
if (root == null) return result;
Stack<TreeNode> s = new Stack<>();
//Stack<TreeNode> out = new Stack<>();
s.push(root);
while ( ! s.isEmpty()) {
TreeNode currNode = s.pop();
//if (currNode != null) out.push(currNode);
if (currNode != null) result.add(0, currNode.val);
if (currNode.left != null) s.push(currNode.left);
if (currNode.right != null) s.push(currNode.right);
}
/*
while ( ! out.isEmpty()) {
TreeNode outNode = out.pop();
result.add(outNode.val);
}
*/
return result;
}
// recursive pseudocode
/*
postOrder(node)
if (node == null)
return
postOrder(node.left)
postOrder(node.right)
visit(node)
*/
}