94. Binary Tree Inorder Traversal (M)

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?

题意

求树的中序遍历。

思路

递归或者迭代。

另有一种神奇的既不使用栈也不用递归的遍历方法，参考 Inorder Tree Traversal without recursion and without stack!。(官方解答改变了树的结构却没有恢复原状)

代码实现 - 递归

/**
 * 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> ans = new ArrayList<>();
        inorder(root, ans);
        return ans;
    }
    private void inorder(TreeNode x, List<Integer> ans) {
        if (x == null) {
            return;
        }
        inorder(x.left, ans);
        ans.add(x.val);
        inorder(x.right, ans);
    }
}

代码实现 - 迭代

/**
 * 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> ans = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            ans.add(cur.val);
            cur = cur.right;
        }
        return ans;
    }
}

代码实现 - O(1)空间迭代

/**
 * 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> ans = new ArrayList<>();
        TreeNode cur = root;
        while (cur != null) {
            if (cur.left != null) {
                TreeNode r = cur.left;
                while (r.right != null && r.right != cur) {
                    r = r.right;
                }
                if (r.right == null) {
                    r.right = cur;
                    cur = cur.left;
                } else {
                    r.right = null;
                    ans.add(cur.val);
                    cur = cur.right;
                }
            } else {
                ans.add(cur.val);
                cur = cur.right;
            }
        }
        return ans;
    }
}