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  • 298. Binary Tree Longest Consecutive Sequence

    题目:

    Given a binary tree, find the length of the longest consecutive sequence path.

    The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from parent to child (cannot be the reverse).

    For example,

       1
        
         3
        / 
       2   4
            
             5
    

    Longest consecutive sequence path is 3-4-5, so return 3.

       2
        
         3
        / 
       2    
      / 
     1
    

    Longest consecutive sequence path is 2-3,not3-2-1, so return 2.

    链接: http://leetcode.com/problems/binary-tree-longest-consecutive-sequence/

    题解:

    求二叉树中最长连续序列。 题目又比较长, 不过我们可以确定这个最长连续序列肯定是递增的,比如123,或者345。知道这点以后就可以用DFS遍历了。

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

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        private int max = 0;
        public int longestConsecutive(TreeNode root) {
            if(root == null) {
                return max;
            }
            findLongestConsecutive(root, 0, root.val);
            return max;
        }
        
        private void findLongestConsecutive(TreeNode root, int curMax, int target) {
            if(root == null) {
                return;
            }
            if(root.val == target) {
                curMax++;
            } else {
                curMax = 1;
            }
            max = Math.max(max, curMax);
            findLongestConsecutive(root.left, curMax, root.val + 1);
            findLongestConsecutive(root.right, curMax, root.val + 1);
        }
    }

    二刷:

    根据题目的意思,最长连续子序列必须是从root到leaf的方向。 比如 1->2,那么我们就返回长度2, 比如1->3->4->5,我们就返回3->4->5这个子序列的长度3。把树遍历一遍就可以得到结果了。

    方法和一刷一样,构建一个辅助方法,形参是当前要处理的节点root,root父节点的值lastVal,以及当前的长度curLen。也要使用一个global variable maxLen来保存全局最大长度。也有使用stack来iterative遍历的做法。

    Java:

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

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        private int maxLen = 0;
        
        public int longestConsecutive(TreeNode root) {
            longestConsecutive(root, 0, 0);
            return maxLen;
        }
        
        private void longestConsecutive(TreeNode root, int lastVal, int curLen) {
            if (root == null) return;
            if (root.val != lastVal + 1) curLen = 1;
            else curLen++;
            maxLen = Math.max(maxLen, curLen);
            longestConsecutive(root.left, root.val, curLen);
            longestConsecutive(root.right, root.val, curLen);
        }
    }

    Reference:

    https://leetcode.com/discuss/68723/simple-recursive-dfs-without-global-variable

    https://leetcode.com/discuss/66486/c-solution-in-4-lines

    https://leetcode.com/discuss/66565/1ms-easy-understand-java-solution-just-traverse-the-tree-once

    https://leetcode.com/discuss/68094/dont-understand-what-is-consecutive-sequence

    https://leetcode.com/discuss/66646/two-simple-iterative-solutions-bfs-and-dfs

    https://leetcode.com/discuss/66548/recursive-solution-bottom-iteration-solution-using-stack

    https://leetcode.com/discuss/66584/easy-java-dfs-is-there-better-time-complexity-solution

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/5047038.html
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