迭代实现后序遍历比迭代实现先序和中序都要稍微复杂，对其一直有恐惧心理，总觉得自己不看答案做不出来……这次竟然做出来了，很开心，练习还是有效果的~
思路1：遍历到curr节点，不管三七二十一，入栈，继续访问左节点，因为后序遍历中root最后访问。问题是栈中的节点何时出栈？答案是其左右节点均被访问之后。怎么判断其左右节点均被访问过了呢？或者换个问题，怎么判断右节点没有被访问过（一路向左，左节点必然先被访问了）？答案是如果栈顶元素有右节点且右节点没有出现在结果列表的最后。
思路2：先序遍历是root->left->right,后序遍历是left->right->root,因此修改下先序遍历实现，让其访问顺序变为root->right->left，则逆序所得结果即为后序遍历结果。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
// Method 2: modification of preorder from root->left->right to root->right->left
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> postorder = new ArrayList<Integer>();
if (root == null) return postorder;
LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
if (curr != null) {
postorder.add(curr.val);
if (curr.left != null)
stack.push(curr.left);
curr = curr.right;
} else {
curr = stack.pop();
}
}
Collections.reverse(postorder);
return postorder;
}
// Method 1
public List<Integer> postorderTraversal1(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
if (root == null) return ret;
LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
if (curr != null) {
stack.push(curr);
curr = curr.left;
} else {
TreeNode top = stack.peek();
if (top.right != null && (ret.size() == 0 || ret.get(ret.size() - 1) != top.right.val))
curr = top.right; // right child has not been visited
else // no right child or has been visited
ret.add(stack.pop().val);
}
}
return ret;
}
}