/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
// eg:[10,5,30,-2,6,12,40,null,2,null,null,null,null,null,null,-1]
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
LinkedList<Integer> res = new LinkedList<Integer>();
Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
TreeNode cur = root;
while( cur != null || !stack.isEmpty()){
while(cur != null) {
res.addFirst(cur.val);
stack.push(cur);
cur = cur.right;
}
cur = stack.pop();
cur = cur.left;
}
return res;
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
// eg:[10,5,30,-2,6,12,40,null,2,null,null,null,null,null,null,-1]
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
helper(root, res);
return res;
}
public void helper(TreeNode root, List<Integer> res){
if ( root == null) return;
helper( root.left, res);
helper( root.right, res);
res.add( root.val);
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
// eg:[10,5,30,-2,6,12,40,null,2,null,null,null,null,null,null,-1]
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
TreeNode cur = new TreeNode(-1);
cur.left = root;
while( cur != null){
if (cur.left == null){
cur = cur.right;
} else {
TreeNode temp = cur.left;
while( temp.right != null && temp.right != cur){
temp = temp.right;
}
if( temp.right == null){
temp.right = cur;
cur = cur.left;
} else {
addReverseOrder(cur.left, temp, res);
temp.right = null;
cur = cur.right;
}
}
}
return res;
}
public void reverseOrder(TreeNode start, TreeNode end){
if (start == end ) return;
TreeNode x = start;
TreeNode y = x.right;
while(true){
TreeNode temp = y.right;
y.right = x;
x = y;
y = temp;
if (x == end) break;
}
}
public void addReverseOrder(TreeNode cur, TreeNode pre, List<Integer> res){
reverseOrder(cur, pre);
TreeNode temp = pre;
while(true){
res.add(temp.val);
if(temp == cur) break;
temp = temp.right;
}
reverseOrder(pre,cur);
}
}
/*
1.先建立一个临时结点dummy,并令其左孩子为根结点root,将当前结点设置为dummy;
2.如果当前结点的左孩子为空,则将其右孩子作为当前结点;
3.如果当前结点的左孩子不为空,则找到其在中序遍历中的前驱结点
3.1如果前驱结点的右孩子为空,将它的右孩子设置为当前结点,将当前结点更新为当前结点的左孩子;
3.2如果前驱结点的右孩子为当前结点,倒序输出从当前结点的左孩子到该前驱结点这条路径上所有的结点。将前驱结点的右孩子设置为空,将当前结点更新为当前结点的右孩子。
4.重复以上过程,直到当前结点为空。
*/