Leetcode Binary Tree Longest Consecutive Sequence
2016-06-22 09:18
501 查看
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,
Longest consecutive sequence path is
so return
Longest consecutive sequence path is
so return
Difficulty: Medium
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int ans = 1;
public void helper(TreeNode root, int dep, int pre){
if(root == null)
return;
if(root.val == pre + 1){
ans = Math.max(ans, dep + 1);
helper(root.left, dep + 1, root.val);
helper(root.right, dep + 1, root.val);
}
else{
helper(root.left, 1, root.val);
helper(root.right, 1, root.val);
}
return;
}
public int longestConsecutive(TreeNode root) {
if(root == null)
return 0;
helper(root.left, 1, root.val);
helper(root.right, 1, root.val);
return ans;
}
}
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,not
3-2-1,
so return
2.
Difficulty: Medium
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int ans = 1;
public void helper(TreeNode root, int dep, int pre){
if(root == null)
return;
if(root.val == pre + 1){
ans = Math.max(ans, dep + 1);
helper(root.left, dep + 1, root.val);
helper(root.right, dep + 1, root.val);
}
else{
helper(root.left, 1, root.val);
helper(root.right, 1, root.val);
}
return;
}
public int longestConsecutive(TreeNode root) {
if(root == null)
return 0;
helper(root.left, 1, root.val);
helper(root.right, 1, root.val);
return ans;
}
}
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