Created
July 23, 2015 12:38
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Three way of post order traversal
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| /** | |
| * Definition for a binary tree node. | |
| * public class TreeNode { | |
| * int val; | |
| * TreeNode left; | |
| * TreeNode right; | |
| * TreeNode(int x) { val = x; } | |
| * } | |
| */ | |
| public class Solution { | |
| public List<Integer> postorderTraversal(TreeNode root) { | |
| List<Integer> l = new LinkedList<Integer>(); | |
| if(root == null) { | |
| return l; | |
| } | |
| Stack<TreeNode> s1 = new Stack<>(); | |
| Stack<TreeNode> s2 = new Stack<>(); | |
| s1.push(root); | |
| while(!s1.isEmpty()) { | |
| TreeNode n = s1.pop(); | |
| s2.push(n); | |
| if(n.left != null) { | |
| s1.push(n.left); | |
| } | |
| if(n.right != null) { | |
| s1.push(n.right); | |
| } | |
| } | |
| while(!s2.isEmpty()) { | |
| l.add(s2.pop().val); | |
| } | |
| return l; | |
| } | |
| public List<Integer> postorderTraversalOneStack(TreeNode root) { | |
| List<Integer> l = new LinkedList<Integer>(); | |
| Stack<TreeNode> s = new Stack<>(); | |
| TreeNode traverse = root; | |
| TreeNode prev = null; | |
| while(traverse != null || !s.isEmpty()) { | |
| while(traverse != null) { | |
| s.push(traverse); | |
| traverse = traverse.left; | |
| } | |
| if(s.peek().right != null && prev != s.peek().right) { | |
| traverse = s.peek().right; | |
| } else { | |
| l.add(s.peek().val); | |
| prev = s.pop(); | |
| } | |
| } | |
| return l; | |
| } | |
| public List<Integer> postOrderRecursive(List<Integer> l, TreeNode node) { | |
| if(node == null) { | |
| return l; | |
| } | |
| postOrderRecursive(l, node.left); | |
| postOrderRecursive(l, node.right); | |
| l.add(node.val); | |
| return l; | |
| } | |
| } |
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