117. Populating Next Right Pointers in Each Node II

題目：

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

 

 

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  Tree Depth-first Search  

連接： http://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

題解：

同樣是DFS。跟上題不一樣在於，給定輸入不是徹底二叉樹了，因此要加入一些條件來判斷是否存在一個合理的next值。而且對左節點和右節點的有效性也要驗證。最後要先遞歸鏈接右節點，再connect左節點。

Time Complexity - O(n)， Space Complexity - O(1)。

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

Update:

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null)
            return;
        TreeLinkNode p = root.next;
        
        while(p != null) {
            if(p.left != null) {
                p = p.left;
                break;
            } else if (p.right != null) {
                p = p.right;
                break;
            }
            p = p.next;    
        }
        
        if(root.right != null) 
            root.right.next = p;
        if(root.left != null) 
            root.left.next = (root.right != null) ? root.right : p;
        
        connect(root.right);
        connect(root.left);
    }
}

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二刷:

依然使用了遞歸，並無作到constant space。留給三刷了。

Java:

Time Complexity - O(n)， Space Complexity - O(n)。

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

 

iterative:

Level order traversal。主要就是相似二叉樹層序遍歷。這回把頂層看做一個linkedlist，咱們只須要繼續鏈接這linkedlist中每一個節點的子節點們。當頂層遍歷完畢之後，下一層正好也造成了一個新的類linkedlist。咱們換到下一層之後繼續遍歷，直到最後。

 

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode curLevel = new TreeLinkNode(-1);
        TreeLinkNode newLevel = curLevel;
        while (root != null) {
            if (root.left != null) {
                curLevel.next = root.left;
                curLevel = curLevel.next;
            }
            if (root.right != null) {
                curLevel.next = root.right;
                curLevel = curLevel.next;
            }
            root = root.next;
            if (root == null) {
                curLevel = newLevel;
                root = newLevel.next;
                newLevel.next = null;
            }
        }
    }
}

 

Update:

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) return;
        TreeLinkNode curLevel = new TreeLinkNode(-1);
        TreeLinkNode newLevel = curLevel;
        while (root != null) {
            if (root.left != null) {
                curLevel.next = root.left;
                curLevel = curLevel.next;
            }
            if (root.right != null) {
                curLevel.next = root.right;
                curLevel = curLevel.next;
            }
            root = root.next;
            if (root == null) {
                root = newLevel.next;
                newLevel.next = null;
                curLevel = newLevel;
            }
         }
        
    }
}

 

 

Reference:

http://www.cnblogs.com/springfor/p/3889327.html

https://leetcode.com/discuss/67291/java-solution-with-constant-space

https://leetcode.com/discuss/60795/o-1-space-o-n-time-java-solution

https://leetcode.com/discuss/24398/simple-solution-using-constant-space

https://leetcode.com/discuss/65526/ac-python-o-1-space-solution-12-lines-and-easy-to-understand

https://leetcode.com/discuss/3339/o-1-space-o-n-complexity-iterative-solution