124. Binary Tree Maximum Path Sum

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node’s values in the path.

Given the root of a binary tree, return the maximum path sum of any path.

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Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

ex1

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Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

ex2

Constraints:

  • The number of nodes in the tree is in the range [1, 3 * 104].
  • -1000 <= Node.val <= 1000

Solution

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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 {
    int maxSum = 0;
    public int maxPathSum(TreeNode root) {
       maxSum = root.val;
       dfs(root);
       return maxSum;
    }
    
    int dfs(TreeNode node) {
        if (node == null) return 0;
        
        int leftSum = Math.max(dfs(node.left), 0);
         // int leftSum = dfs(node.left);
        int rightSum = Math.max(dfs(node.right), 0);
        // int rightSum = dfs(node.right);
        
        int currSum = leftSum + node.val + rightSum;
        
        maxSum = Math.max(currSum, maxSum);
        
        return node.val + Math.max(leftSum, rightSum);
    }
}

Solution 2021-11-22

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class Solution {
    int max;
    public int maxPathSum(TreeNode root) {
        if (root == null) return 0;
        max = root.val;
        maxSum(root);
        return max;
    }
    
    int maxSum(TreeNode node) {
        if (node == null) return 0;
        int leftSum = Math.max(maxSum(node.left), 0); // either take left or not
        int rightSum = Math.max(maxSum(node.right), 0); // either take right or not
        int currMax = leftSum + rightSum + node.val;
        max = Math.max(max, currMax);
        return node.val + Math.max(leftSum, rightSum); 
    }
}

Solution 2022-01-30

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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 {
    int maxSum = 0;
    public int maxPathSum(TreeNode root) {
        maxSum = root.val;
        dfs(root);
        return maxSum;
    }

    int dfs(TreeNode node) {
        if (node == null) return 0;
        
        int left = Math.max(dfs(node.left), 0);
        int right = Math.max(dfs(node.right), 0);
        
        int currSum = node.val + left + right;
        maxSum = Math.max(maxSum, currSum);
        
        return node.val + Math.max(left, right);
    }
}