Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree [1,null,2,3]
,
1 2 / 3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
在做99. Recover Binary Search Tree 时,要求不用O(n)的空间复杂度,只用constant space,因此不能用普通的迭代和递归法,递归里有函数栈,迭代里有用户声明的栈,都会带来O(lgn)-O(n)的空间复杂度。
用线索二叉树。
在迭代中,栈里存的都是根节点,是为了遍历完左子树后能重新找到根,再遍历根和右子树。想一种办法,能不用栈又能找回根节点。办法是:对每个根节点,首先找到在左子树里的它的前继节点,前继节点的right指针是闲置的,把闲置的right指针指向根,就可以重新找到根了。
什么时候恢复right指针、并且在遍历这个根节点的时候如何知道,已经遍历过这个根节点的左子树,还是要开始遍历整个根下的树?
在一步步找根的前继节点时,如果找到right == root,说明之前已经重置过前继节点的right指针了,这次应该恢复right指针,并遍历root的右子树;如果找到right == null,说明之前未遍历过root树,这次应当重置right指针,并开始遍历root的左子树。
/** * 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) { TreeNode cur = root, prev = null; List<Integer> re = new LinkedList<Integer>(); while (cur != null) { prev = cur.left; if (prev == null) { re.add(cur.val); // leaf node, traverse current node cur = cur.right; // jump to its next continue; } while (true) { if (prev.right == null) { prev.right = cur; cur = cur.left; break; } else if (prev.right == cur) { prev.right = null; re.add(cur.val); // root node, traverse current node cur = cur.right; break; } prev = prev.right; } } return re; } }
普通的迭代法:
/** * 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> re = new LinkedList<Integer>(); Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode tmp = root, right; while (tmp != null) { stack.push(tmp); tmp = tmp.left; } while (!stack.isEmpty()) { tmp = stack.pop(); re.add(tmp.val); right = tmp.right; while (right != null) { stack.push(right); right = right.left; } } return re; } }
时间: 2024-11-03 05:23:10