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Max_Path_Sum.java
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Max_Path_Sum.java
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/*
The secret for this problem is the post-order traversal: you want to process a node's right and left subtree before you process the node itself. With the data from a node's right and left subtree you can decide what to return: either:
the value of the node (both subtrees yield negative value) -- even if the node itself is negative, as we need to satisfy the requirement that a path must have at least one node,
the value of the node + the value from its left or right subtree, whichever is positive (one of the subtrees yield negative value)
the value of the node + the value of both subtrees. (if both subtrees yield a positive value)
.
Runtime: O(n)
Space Complexity for balanced tree: O(log n). Worst case (tree degenerated into a linked list) O(n);
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxPathSum(TreeNode root) {
int [] res = new int[1];
res[0] = Integer.MIN_VALUE;
maxPathSum(root, res);
return res[0];
}
public int maxPathSum(TreeNode root, int[] max) {
if (root == null) return -1;
int left = maxPathSum(root.left, max);
int right = maxPathSum(root.right, max);
int curSum = root.val + Math.max(left, 0) + Math.max(right, 0);
max[0] = Math.max(curSum, max[0]);
return root.val + Math.max(Math.max(left, right),0);
}
}