LeetCode - Binary Tree Maximum Path Sum
Problem description
Given a non-empty binary tree, find the maximum path sum.
For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.
Example 1:
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Input: [1,2,3]
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/ \
2 3
Output: 6
Example 2:
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Input: [-10,9,20,null,null,15,7]
-10
/ \
9 20
/ \
15 7
Output: 42
Analysis
The general idea is to analysis for each node, there are two possible path. One is with its parent to form a path, another one is don’t with it’s parent. So we can use a max to store those without it’s parent and return the other value to it’s parent so that we can do the recursion.
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxPathSum(TreeNode root) {
int v = dfs(root);
return Math.max(max, v);
}
int max = Integer.MIN_VALUE;
public int dfs(TreeNode node){
if (node == null){
return 0;
}
if (node.left == null && node.right == null){
max = Math.max(node.val, max);
return node.val;
}
else if (node.left == null){
int right = Math.max(0, dfs(node.right));
max = Math.max(max, right+node.val);
return right + node.val;
}
else if (node.right == null){
int left = Math.max(0, dfs(node.left));
max = Math.max(max, left+node.val);
return left + node.val;
}
else {
int left = Math.max(0, dfs(node.left));
int right = Math.max(0, dfs(node.right));
int valNotEndAtNode = Math.max(left, right) + node.val;
int valEndAtNode = left + right + node.val;
max = Math.max(valNotEndAtNode, Math.max(max, valEndAtNode));
return valNotEndAtNode;
}
}
}
What to improve
- Note that there’s a possible path which is it only contains node itself.