Binary Tree Postorder Traversal

Change-Id: I14aeed7075fb90bace971602710e62155f2136d6

Author: applewjg

BinaryTreePostorderTraversal.java

@@ -0,0 +1,104 @@
+/*
+ Author:     King, [email protected]
+ Date:       Nov 20, 2014
+ Problem:    Binary Tree Postorder Traversal 
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/binary-tree-postorder-traversal/
+ Notes:
+ Given a binary tree, return the postorder traversal of its nodes' values.
+
+ For example:
+ Given binary tree {1,#,2,3},
+    1
+     \
+      2
+     /
+    3
+ return [3,2,1].
+
+ Note: Recursive solution is trivial, could you do it iteratively?
+
+ Solution: 1. Iterative way (stack).   Time: O(n), Space: O(n).
+           2. Recursive solution.      Time: O(n), Space: O(n).
+           3. Threaded tree (Morris).  Time: O(n), Space: O(n/1). 
+              Space: O(1) if in-place reverse.
+              You may refer to my blog for more detailed explanations: 
+              http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html
+*/
+
+
+/**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public List<Integer> postorderTraversal_1(TreeNode root) {
+        List<Integer> res = new ArrayList<Integer>();
+        if (root == null) return res;
+        Stack<TreeNode> stk = new Stack<TreeNode>();
+        TreeNode cur = root;
+        TreeNode pre = null;
+        while (stk.isEmpty() == false || cur != null) {
+            if (cur != null) {
+                stk.push(cur);
+                cur = cur.left;
+            } else {
+                TreeNode peak = stk.peek();
+                if (peak.right != null && pre != peak.right) {
+                    cur = peak.right;
+                } else {
+                    res.add(peak.val);
+                    stk.pop();
+                    pre = peak;
+                }
+            }
+        }
+        return res;
+    }
+    public List<Integer> postorderTraversal_2(TreeNode root) {
+        List<Integer> res = new ArrayList<Integer>();
+        if (root == null) return res;
+        List<Integer> left = postorderTraversal(root.left);
+        List<Integer> right = postorderTraversal(root.right);
+        res.addAll(left);
+        res.addAll(right);
+        res.add(root.val);
+        return res;
+    }
+    public List<Integer> postorderTraversal_3(TreeNode root) {
+        List<Integer> res = new ArrayList<Integer>();
+        if (root == null) return res;
+        Stack<Integer> stk = new Stack<Integer>();
+        TreeNode dummy = new TreeNode(-1);
+        dummy.left = root;
+        TreeNode cur = dummy;
+        while (cur != null) {
+            if (cur.left == null) {
+                cur = cur.right;
+            } else {
+                TreeNode node = cur.left;
+                while (node.right != null && node.right != cur)
+                    node = node.right;
+                if (node.right == null) {
+                    node.right = cur;
+                    cur = cur.left;
+                } else {
+                    TreeNode temp = cur.left;
+                    while (temp != cur) {
+                        stk.push(temp.val);
+                        temp = temp.right;
+                    }
+                    while (stk.isEmpty() == false) res.add(stk.pop());
+                    node.right = null;
+                    cur = cur.right;
+                }
+            }
+        }
+        return res;
+    }
+}