94 Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

用栈或者Morris遍历

Morris:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        TreeNode prev = null;
        TreeNode curr = root;
        
        while (curr != null) {
            if (curr.left == null) {
                res.add(curr.val);
                prev = curr;
                curr = curr.right;
            } else {
                TreeNode node = curr.left;
                while (node.right != null && node.right != curr)
                    node = node.right;
                
                if (node.right == null) {
                    node.right = curr;
                    curr = curr.left;
                } else {
                    res.add(curr.val);
                    prev = curr;
                    curr = curr.right;
                    node.right = null;
                }
            }
        }
        return res;
    }
}

Stack:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> s = new Stack<>();
        TreeNode p = root;
        
        while (p != null || !s.isEmpty()) {
            if (p != null) {
                s.push(p);
                p = p.left;
            } else {
                p = s.pop();
                res.add(p.val);
                p = p.right;
            }
        }
        return res;
    }
}