298. Binary Tree Longest Consecutive Sequence (Medium)

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.

Solution 1:

version 1: 59ms

这道题让我们求二叉树的最长连续序列，关于二叉树的题基本都需要遍历树，而递归遍历写起来特别简单，下面这种解法是用到了递归版的先序遍历，我们对于每个遍历到的节点，我们看节点值是否比参数值(父节点值)大1，如果是则长度加1，否则长度重置为1，然后更新结果res，再递归调用左右子节点即可，参见代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    void preOrderTraverse(TreeNode* root, int pre, int len, int& res) {
        if (!root) return;
        if (root->val == pre+1) ++len;
        else len = 1;
        res = max(res, len);
        preOrderTraverse(root->left, root->val, len, res);
        preOrderTraverse(root->right, root->val, len, res);
    }
public:
    int longestConsecutive(TreeNode* root) {
        if (!root) return 0;
        int res = 0;
        preOrderTraverse(root, root->val-1, 0, res);
        return res;
    }
};

version 2: 36ms

下面这种写法是利用分治法的思想，对左右子节点分别处理，如果左子节点存在且节点值比其父节点值大1，则递归调用函数，如果节点值不是刚好大1，则递归调用重置了长度的函数，对于右子节点的处理情况和左子节点相同，参见代码如下：

class Solution {
    void preOrderTraverse(TreeNode* root, int len, int& res) {
        res = max(res, len);
        if (root->left) {
            if (root->left->val == root->val+1)
                preOrderTraverse(root->left, len+1, res);
            else 
                preOrderTraverse(root->left, 1, res);
        }
        if (root->right) {
            if (root->right->val == root->val+1)
                preOrderTraverse(root->right, len+1, res);
            else 
                preOrderTraverse(root->right, 1, res);
        }
    }

public:
    int longestConsecutive(TreeNode* root) {
        if (!root) return 0;
        int res = 1;
        preOrderTraverse(root, 1, res);
        return res;
    }
};

version 3: 29ms

下面这种递归写法相当简洁，但是核心思想和上面两种方法并没有太大的区别，参见代码如下：

class Solution {
    int preOrderTraverse(TreeNode* root, TreeNode* parent, int len) {
        if (!root) return len;
        len = (parent && root->val == parent->val+1) ? len+1:1;
        return max(len, max(preOrderTraverse(root->left, root, len), preOrderTraverse(root->right, root, len)));
    }

public:
    int longestConsecutive(TreeNode* root) {
        return preOrderTraverse(root, NULL, 0);
    }
};

Solution 2: DFS