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Populating Next Right Pointers in Each Node
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Populating Next Right Pointers in Each Node
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1 / 2 3 / \ / 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / 2 -> 3 -> NULL / \ / 4-> 5->6 -> 7 -> NULL
此题,我的第一反映是广度优先遍历。按层访问。每一层自左向右按顺序链接!需要借助于队列来实现。但是没有去实现。
思路二是,我们仔细观察发现,链接分为2种情况:情况1是链接root节点的左子树与右子树。情况2是链接节点2的右子树与节点3的左子树。
1 2 ——》 3 / \ / \ / 2 -> 3 4 5 ——》 6 7
(1) (2)
剩余的就是选择一个普通的遍历方式。
public static void tra(TreeLinkNode root) { if (root == null) { return; } Stack<TreeLinkNode> stack = new Stack<>(); while (null != root || !stack.isEmpty()) { while (null != root) { stack.push(root);
//visit
System.out.println(root.val);
// root = root.left; } root = stack.pop(); root = root.right; } }
遍历的同时,链接2种。
public static void connect(TreeLinkNode root) { if (root == null) { return; } Stack<TreeLinkNode> stack = new Stack<>(); while (null != root || !stack.isEmpty()) { while (null != root) { stack.push(root); if (root.left != null) { root.left.next = root.right; if (root.next != null) { root.right.next = root.next.left; } } root = root.left; } root = stack.pop(); root = root.right; } }
Populating Next Right Pointers in Each Node
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