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问题描述
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:
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
解决思路
层次遍历。
开始节点为根节点，下一层为节点的左子结点。
连接的过程中注意提前判断节点非空。
程序
public class Solution {
public void connect(TreeLinkNode root) {
if (root == null) {
return ;
}
TreeLinkNode curLevel = root;
while (curLevel != null) {
TreeLinkNode cur = curLevel;
while (cur != null) {
TreeLinkNode next = cur.next;
if (cur.left != null && cur.right != null) {
cur.left.next = cur.right;
}
if (cur.right != null && next != null) {
cur.right.next = next.left;
}
cur = next;
}
curLevel = curLevel.left;
}
}
}
Follow up
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
For example,
Given the following binary tree,
1
/ 2 3
/ \ 4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ 2 -> 3 -> NULL
/ \ 4-> 5 -> 7 -> NULL
程序
public class Solution {
public void connect(TreeLinkNode root) {
if (root == null) {
return ;
}
TreeLinkNode curLevel = root;
while (curLevel != null) {
TreeLinkNode cur = curLevel;
TreeLinkNode nextLevel = null; // mark the next node
while (cur != null) {
TreeLinkNode next = cur.next;
while (next != null && next.left == null && next.right == null) {
next = next.next; // skip node with no children
}
if (nextLevel == null) {
// mark the first no-empty node as the next node
if (cur.left != null) {
nextLevel = cur.left;
} else if (cur.right != null) {
nextLevel = cur.right;
}
}
if (cur.left != null && cur.right != null) {
cur.left.next = cur.right;
}
if (next != null) {
// connect cross the subtrees
if (cur.right != null) {
cur.right.next = next.left == null ? next.right : next.left;
} else if (cur.left != null) {
cur.left.next = next.left == null ? next.right : next.left;
}
}
cur = next;
}
curLevel = nextLevel;
}
}
}
Populating Next Right Pointers in Each Node
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原文地址：http://www.cnblogs.com/harrygogo/p/4655295.html
