LeetCode - Populating Next Right Pointers in Each Node II 及其变形题

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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

第一个题的要求是完全二叉树，第二题没有这个要求。很显然如果没有空间复杂度的要求，O(1)，这题完全可以按层次遍历的思想搞，又提到了工具方法的概念，可以先看这道题：http://blog.csdn.net/my_jobs/article/details/47665089按层次遍历。完全可以改造这个题而成。但是这两个题都要求空间复杂度为O(1)。所以此办法不行，但是可以给出代码，对这两个题完全通用：

    public void connect(TreeLinkNode root) {
        if (root == null) return;
        LinkedList<TreeLinkNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int len = queue.size();
            TreeLinkNode pre = null;
            for (int i = 0; i < len; i++) {
                TreeLinkNode node = queue.poll();
                if (i == 0) {
                    pre = node;
                } else {
                    pre.next = node;
                    pre = pre.next;
                }
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
        }
    }

先做第一个，比较好做。看代码就明白了：

    public void connect(TreeLinkNode root) {
        if (root == null) return;
        root.next = null;
        while (root.left != null) {
            TreeLinkNode p = root;
            while (p != null) {
                p.left.next = p.right;
                if (p.next != null) {
                    p.right.next = p.next.left;
                } else {
                    p.right.next = null;
                }
                p = p.next;
            }
            root = root.next;
        }
    }

在看第二道题，第二题就难了。

    public void connect(TreeLinkNode root) {
        while(root != null){
            TreeLinkNode childHead = new TreeLinkNode(0);
            TreeLinkNode p = childHead;
            while (root != null){
                if (root.left != null) { 
                    p.next = root.left; 
                    p = p.next;
                }
                if (root.right != null) { 
                    p.next = root.right; 
                    p = p.next;
                }
                root = root.next;
            }
            p.next = null;
            root = childHead.next;
        }
    }

改造题：

Given the following binary tree,

         1
       /        2    3
       \  /        5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /   
      2 -> 3 -> NULL
     /       
    5->6->7 -> NULL

    public TreeNode treeToList(TreeNode root) {
    	if (root == null) return null;
    	LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
    	queue.add(root);
    	TreeNode head = new TreeNode(0);
    	while (!queue.isEmpty()) {
    		int len = queue.size();
    		TreeNode pre = null;
    		for (int i = 0; i < len; i++) {
    			TreeNode node = queue.poll();
    			if (node.left != null) queue.add(node.left);
    			if (node.right != null) queue.add(node.right);
    			node.left = null;
    			node.right = null;
    			if (i == 0) {
    				pre = node;
    				head.left = pre;
    				head = head.left;
    			} else {
    				pre.right = node;
    				pre = pre.right;
    			}
    		}
    	}
    	return root;
    }

LeetCode - Populating Next Right Pointers in Each Node II 及其变形题

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原文地址：http://blog.csdn.net/my_jobs/article/details/47683953