Monday, April 17, 2017

LeetCode 549 - Binary Tree Longest Consecutive Sequence II


LeetCode 298 - Binary Tree Longest Consecutive Sequence
http://bookshadow.com/weblog/2017/04/09/leetcode-binary-tree-longest-consecutive-sequence-ii/
Given a binary tree, you need to find the length of Longest Consecutive Path in Binary Tree.
Especially, this path can be either increasing or decreasing. For example, [1,2,3,4] and [4,3,2,1] are both considered valid, but the path [1,2,4,3] is not valid. On the other hand, the path can be in the child-Parent-child order, where not necessarily be parent-child order.
Example 1:
Input:
        1
       / \
      2   3
Output: 2
Explanation: The longest consecutive path is [1, 2] or [2, 1].
Example 2:
Input:
        2
       / \
      1   3
Output: 3
Explanation: The longest consecutive path is [1, 2, 3] or [3, 2, 1].

解法I 一趟遍历
时间复杂度O(n) n为节点的个数
定义函数solve(root),递归求解以root为根节点向子节点方向(parent-child)的路径中,最大连续递增路径长度inc,以及最大连续递减路径长度dec

则以root为根节点的子树中,最大连续路径长度=inc + dec + 1(路径不包含root)
https://discuss.leetcode.com/topic/85764/neat-java-solution-single-pass-o-n
    int maxval = 0;
    public int longestConsecutive(TreeNode root) {
        longestPath(root);
        return maxval;
    }
    public int[] longestPath(TreeNode root) {
        if (root == null)
            return new int[] {0,0};
        int inr = 1, dcr = 1;
        if (root.left != null) {
            int[] l = longestPath(root.left);
            if (root.val == root.left.val + 1)
                dcr = l[1] + 1;
            else if (root.val == root.left.val - 1)
                inr = l[0] + 1;
        }
        if (root.right != null) {
            int[] r = longestPath(root.right);
            if (root.val == root.right.val + 1)
                dcr = Math.max(dcr, r[1] + 1);
            else if (root.val == root.right.val - 1)
                inr = Math.max(inr, r[0] + 1);
        }
        maxval = Math.max(maxval, dcr + inr - 1);
        return new int[] {inr, dcr};
    }
https://discuss.leetcode.com/topic/85745/java-solution-binary-tree-post-order-traversal
    int max = 0;
    
    class Result {
        TreeNode node;
        int inc;
        int des;
    }
    
    public int longestConsecutive(TreeNode root) {
        traverse(root);
        return max;
    }
    
    private Result traverse(TreeNode node) {
        if (node == null) return null;
        
        Result left = traverse(node.left);
        Result right = traverse(node.right);
        
        Result curr = new Result();
        curr.node = node;
        curr.inc = 1;
        curr.des = 1;
        
        if (left != null) {
            if (node.val - left.node.val == 1) {
                curr.inc = Math.max(curr.inc, left.inc + 1);
            }
            else if (node.val - left.node.val == -1) {
                curr.des = Math.max(curr.des, left.des + 1);
            }
        }
        
        if (right != null) {
            if (node.val - right.node.val == 1) {
                curr.inc = Math.max(curr.inc, right.inc + 1);
            }
            else if (node.val - right.node.val == -1) {
                curr.des = Math.max(curr.des, right.des + 1);
            }
        }
        
        max = Math.max(max, curr.inc + curr.des - 1);
        
        return curr;
    }

解法II 递归 + 遍历二叉树
时间复杂度O(n^2) n为节点的个数
定义函数maxLength,递归计算从根节点出发向叶子节点可以得到的最长连续路径长度。

分别记根节点root的左右孩子为lchild, rchild

若lchild与root的差值为1,调用maxLength得到左子路径长度lsize

若rchild与root的差值为1,调用maxLength得到右子路径长度rsize

若lchild < root < rchild或者lchild > root > rchild:

则当前最长路径长度为lsize + rsize + 1

否则,最长路径为max(lsize, rsize) + 1

遍历二叉树,求最大值。
https://xuezhashuati.blogspot.com/2017/04/lintcode-619-binary-tree-longest.html
用分治法。
我们用ResultType来记录从某一个点往下走的时候递增的最大路径和递减的最大路径,以及一个全局的最长路径。
在某一点,全局的最长路径就是:
1:左子树中遇到的最长路径
2:右子树中遇到的最长路径
3:通过当前点的最长路径
这三者的最大值。
    class ResultType {
        int maxLength;
        int maxUp;
        int maxDown;
        public ResultType(int maxLength, int maxUp, int maxDown) {
            this.maxLength = maxLength;
            this.maxUp = maxUp;
            this.maxDown = maxDown;
        }
    }
    public int longestConsecutive2(TreeNode root) {        
        ResultType result = helper(root);
        return result.maxLength;
    }
    
    public ResultType helper(TreeNode root) {
        
        if (root == null) {
            return new ResultType(0, 0, 0);
        }
        
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        
        int up = 0;
        int down = 0;
        
        if (root.left != null && root.left.val + 1 == root.val) {
            down = Math.max(down, left.maxDown + 1);
        }
        
        if (root.left != null && root.left.val - 1 == root.val) {
            up = Math.max(up, left.maxUp + 1);
        }
        
        if (root.right != null && root.right.val + 1 == root.val) {
            down = Math.max(down, right.maxDown + 1);
        }
        
        if (root.right != null && root.right.val - 1 == root.val) {
            up = Math.max(up, right.maxUp + 1);
        }
        
        int max = Math.max(Math.max(left.maxLength, right.maxLength), up + down + 1);
        return new ResultType(max, up, down);
    }

LintCode - 619 Binary Tree Longest Consecutive Sequence III
https://xuezhashuati.blogspot.com/2017/04/lintcode-619-binary-tree-longest_22.html
It's follow up problem for Binary Tree Longest Consecutive Sequence II

Given a k-ary tree, find the length of the longest consecutive sequence path.
The path could be start and end at any node in the tree

Example
An example of test data: k-ary tree 5<6<7<>,5<>,8<>>,4<3<>,5<>,3<>>>, denote the following structure:
     5
   /   \
  6     4
 /|\   /|\
7 5 8 3 5 3
Return 5, // 3-4-5-6-7
思路
和Binary Tree Longest Consecutive Sequence II一样的做法。
II只要check一下left和right。
这题因为有多个子节点,所以要用一个循环来check所有。

    class ResultType {
        int globalMax;
        int maxUp;
        int maxDown;
        public ResultType(int globalMax, int maxUp, int maxDown) {
            this.globalMax = globalMax;
            this.maxUp = maxUp;
            this.maxDown = maxDown;
        }
    }
    
    public int longestConsecutive3(MultiTreeNode root) {        
        return helper(root).globalMax;
    }
    
    public ResultType helper(MultiTreeNode root) {
        
        if (root == null) {
            return new ResultType(0, 0, 0);
        }
        
        int maxUp = 0;
        int maxDown = 0;
        int max = Integer.MIN_VALUE;
        
        for (MultiTreeNode child : root.children) {
            
            if (child == null) {
                continue;
            }
            
            ResultType childResult = helper(child);
            if (child.val + 1 == root.val) {
                maxDown = Math.max(maxDown, childResult.maxDown + 1);
            }
            
            if (child.val - 1 == root.val) {
                maxUp = Math.max(maxUp, childResult.maxUp + 1);
            }
            
            max = Math.max(Math.max(max, childResult.globalMax), maxUp + maxDown + 1);
        }
        
        return new ResultType(max, maxUp, maxDown);
    }


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