https://oj.leetcode.com/problems/binary-tree-inorder-traversal/
http://blog.csdn.net/linhuanmars/article/details/20187257
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> inorderTraversal(TreeNode root)
{
// Solution A:
// return inorderTraversal_Recursion(root);
// Solution B:
// return inorderTraversal_StackA(root);
// Solution C:
// return inorderTraversal_StackB(root);
// Solution D:
return inorderTraversal_NoExtraSpace(root);
}
/////////////////////
// Solution D: No Extra Space
// See http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html
//
// Use last pre.right -> cur
public List<Integer> inorderTraversal_NoExtraSpace(TreeNode root)
{
List<Integer> result = new ArrayList<>();
TreeNode cur = root;
TreeNode pre = null;
while (cur != null)
{
if (cur.left == null)
{
result.add(cur.val);
cur = cur.right;
}
else
{
// Find pre of cur in inorder traversal
// which is the most right child in cur‘s left sub tree
pre = cur.left;
while (pre.right != null && pre.right != cur)
pre = pre.right;
if (pre.right == null)
{
// Mark theLastOne.right -> currentMySelf
pre.right = cur;
cur = cur.left;
}
else
{
// All left been visited.
pre.right = null;
result.add(cur.val);
cur = cur.right;
}
}
}
return result;
}
/////////////////////
// Solution C: Stack B
//
public List<Integer> inorderTraversal_StackB(TreeNode root)
{
List<Integer> result = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
while (root != null || !stack.empty())
{
if (root != null)
{
stack.push(root);
root = root.left;
}
else
{
root = stack.pop();
result.add(root.val);
root = root.right;
}
}
return result;
}
/////////////////////
// Solution B: Stack A
//
public List<Integer> inorderTraversal_StackA(TreeNode root)
{
List<Integer> result = new ArrayList<>();
if (root == null)
return result;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.empty())
{
TreeNode node = stack.peek(); // Note!! not pop
if (node.left != null)
{
stack.push(node.left);
continue;
}
node = stack.pop();
result.add(node.val);
if (node.right != null)
stack.push(node.right);
}
return result;
}
/////////////////////
// Solution A: Recursion
//
public List<Integer> inorderTraversal_Recursion(TreeNode root) {
List<Integer> order = new ArrayList<>();
visit(root, order);
return order;
}
private void visit(TreeNode node, List<Integer> order)
{
if (node == null)
return;
visit(node.left, order);
order.add(node.val);
visit(node.right, order);
}
}[LeetCode]94 Binary Tree Inorder Traversal
原文:http://7371901.blog.51cto.com/7361901/1599363