- Binary Tree Longest Consecutive Sequence
中文English
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).
Example
Example 1:
Input:
1
3
/
2 4
5
Output:3
Explanation:
Longest consecutive sequence path is 3-4-5, so return 3.
Example 2:
Input:
2
3
/
2
/
1
Output:2
Explanation:
Longest consecutive sequence path is 2-3,not 3-2-1, so return 2.
解法1:
递归
代码如下:
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: the root of binary tree
* @return: the length of the longest consecutive sequence path
*/
int longestConsecutive(TreeNode * root) {
helper(root);
return maxLen;
}
private:
int maxLen = 0;
int helper(TreeNode * root) {
if (!root) return 0;
if (!root->left && !root->right) return 1;
int leftNum = 0, rightNum = 0;
if (root->left) {
leftNum = helper(root->left);
if (root->val == root->left->val - 1) {
leftNum++;
maxLen = max(maxLen, leftNum);
} else {
maxLen = max(maxLen, leftNum);
leftNum = 1;
}
}
if (root->right) {
rightNum = helper(root->right);
if (root->val == root->right->val - 1) {
rightNum++;
maxLen = max(maxLen, rightNum);
} else {
maxLen = max(maxLen, rightNum);
rightNum = 1;
}
}
return max(leftNum, rightNum);
}
};
二刷:
class Solution {
public:
/**
* @param root: the root of binary tree
* @return: the length of the longest consecutive sequence path
*/
int longestConsecutive(TreeNode *root) {
if (!root) return 0;
helper(root, 1);
return gMaxPathLen;
}
private:
int gMaxPathLen = 0;
void helper(TreeNode *root, int pathLen) {
if (!root) return;
gMaxPathLen = max(gMaxPathLen, pathLen);
if (root->left && root->val + 1 == root->left->val) helper(root->left, pathLen + 1);
else helper(root->left, 1);
if (root->right && root->val + 1 == root->right->val) helper(root->right, pathLen + 1);
else helper(root->right, 1);
return;
}
};