/**
* 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> inorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
TreeNode cur = root;
while( cur != null || !stack.isEmpty()){
while(cur != null){
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
res.add(cur.val);
cur = cur.right;
}
return res;
}
}
/**
* 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> inorderTraversal(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);
res.add(root.val);
helper(root.right, 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> inorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
TreeNode cur = root;
while( cur != null){
if (cur.left == null) {
res.add(cur.val);
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 { // else if (temp.right == cur)
res.add(cur.val);
temp.right = null;
cur = cur.right;
}
}
}
return res;
}
}
/*
1.如果当前结点pNode的左孩子为空，那么输出该结点，并把该结点的右孩子作为当前结点；
2.如果当前结点pNode的左孩子非空，那么就找出该结点在中序遍历中的前驱结点pPre
2.1.当第一次访问该前驱结点pPre时，其右孩子必定为空，那么就将其右孩子设置为当前结点，以便根据这个指针返回到当前结点pNode中，并将当前结点pNode设置为其左孩子；
2.2.当该前驱结点pPre的右孩子为当前结点，那么就输出当前结点，并把前驱结点的右孩子设置为空（恢复树的结构），将当前结点更新为当前结点的右孩子
重复以上两步，直到当前结点为空。
*/