LeetCode 题解（61）: Populating Next Right Pointers in Each Node II

题目：

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:

• You may only use constant extra space.

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

题解：不完全二叉树。自写不简便解法：

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(!root)
            return;
        TreeLinkNode* temp = root;
        while(temp->next)
            temp = temp->next;
        temp->next = NULL;
        
        TreeLinkNode* newRoot = find(root, NULL);
        TreeLinkNode* preNode = find(root, NULL);
        while(preNode) {
            TreeLinkNode* nextNode = find(root, preNode);
            preNode->next = nextNode;
            preNode = nextNode;
        }
        connect(newRoot);
        return;
    }
    
    TreeLinkNode* find(TreeLinkNode* root, TreeLinkNode* first) {
        TreeLinkNode* result = NULL;
        if(!first) {
            while(root && !result) {
                if(root->left) {
                    result = root->left;
                } else if(root->right) {
                    result = root->right;
                } else {
                    root = root->next;
                }
            }
        } else {
            while(root && !result) {
                if(root->left == first) {
                    if(root->right) {
                        result = root->right;
                    } else {
                        root = root->next;
                        while(root && !result) {
                            if(root->left) {
                                result = root->left;
                            } else if(root->right) {
                                result = root->right;
                            } else {
                                root = root->next;
                            }                            
                        }
                    }
                } else if(root->right == first) {
                    root = root->next;
                    while(root && !result) {
                        if(root->left) {
                            result = root->left;
                        } else if(root->right) {
                            result = root->right;
                        } else {
                            root = root->next;
                        }
                    }
                } else {
                    root = root->next;
                }
            }
        }
        return result;
    }
};

网上学来简便解法：注意一定要先处理右子树再处理左子树。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(!root)
            return;
        
        TreeLinkNode* rootNext = root->next;
        TreeLinkNode* next = NULL;
        
        while(rootNext && !next) {
            if(rootNext->left)
                next = rootNext->left;
            else if(rootNext->right)
                next = rootNext->right;
            else
                rootNext = rootNext->next;
        } 
        
        if(root->right) {
            root->right->next = next;
            if(root->left)
                root->left->next = root->right;
        } else {
            if(root->left)
                root->left->next = next;
        }

        connect(root->right);
        connect(root->left);
    }
};

Java版：

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

Python版：

class Solution:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if root == None:
            return
        
        rootNext = root.next
        next = None
        
        while rootNext != None and next == None:
            if rootNext.left != None:
                next = rootNext.left
            elif rootNext.right != None:
                next = rootNext.right
            else:
                rootNext = rootNext.next
                
        if root.right != None:
            root.right.next = next
            if root.left != None:
                root.left.next = root.right
        else:
            if root.left != None:
                root.left.next = next
                
        self.connect(root.right)
        self.connect(root.left)