Question

124. Binary Tree Maximum Path Sum

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Description

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 non-empty path_.

 

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.

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.

 

Constraints:

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

Solutions

Solution 1: Recursion

When thinking about the classic routine of recursion problems in binary trees, we consider:

1. Termination condition (when to terminate recursion)
2. Recursively process the left and right subtrees
3. Merge the calculation results of the left and right subtrees

For this problem, we design a function $dfs(root)$, which returns the maximum path sum of the binary tree with $root$ as the root node.

The execution logic of the function $dfs(root)$ is as follows:

If $root$ does not exist, then $dfs(root)$ returns $0$;

Otherwise, we recursively calculate the maximum path sum of the left and right subtrees of $root$, denoted as $left$ and $right$. If $left$ is less than $0$, then we set it to $0$, similarly, if $right$ is less than $0$, then we set it to $0$.

Then, we update the answer with $root.val + left + right$. Finally, the function returns $root.val + \max(left, right)$.

In the main function, we call $dfs(root)$ to get the maximum path sum of each node, and the maximum value among them is the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def maxPathSum(self, root: Optional[TreeNode]) -> int:
    def dfs(root: Optional[TreeNode]) -> int:
      if root is None:
        return 0
      left = max(0, dfs(root.left))
      right = max(0, dfs(root.right))
      nonlocal ans
      ans = max(ans, root.val + left + right)
      return root.val + max(left, right)

    ans = -inf
    dfs(root)
    return ans

/**
 * 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 {
  private int ans = -1001;

  public int maxPathSum(TreeNode root) {
    dfs(root);
    return ans;
  }

  private int dfs(TreeNode root) {
    if (root == null) {
      return 0;
    }
    int left = Math.max(0, dfs(root.left));
    int right = Math.max(0, dfs(root.right));
    ans = Math.max(ans, root.val + left + right);
    return root.val + Math.max(left, right);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  int maxPathSum(TreeNode* root) {
    int ans = -1001;
    function<int(TreeNode*)> dfs = [&](TreeNode* root) {
      if (!root) {
        return 0;
      }
      int left = max(0, dfs(root->left));
      int right = max(0, dfs(root->right));
      ans = max(ans, left + right + root->val);
      return root->val + max(left, right);
    };
    dfs(root);
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func maxPathSum(root *TreeNode) int {
  ans := -1001
  var dfs func(*TreeNode) int
  dfs = func(root *TreeNode) int {
    if root == nil {
      return 0
    }
    left := max(0, dfs(root.Left))
    right := max(0, dfs(root.Right))
    ans = max(ans, left+right+root.Val)
    return max(left, right) + root.Val
  }
  dfs(root)
  return ans
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function maxPathSum(root: TreeNode | null): number {
  let ans = -1001;
  const dfs = (root: TreeNode | null): number => {
    if (!root) {
      return 0;
    }
    const left = Math.max(0, dfs(root.left));
    const right = Math.max(0, dfs(root.right));
    ans = Math.max(ans, left + right + root.val);
    return Math.max(left, right) + root.val;
  };
  dfs(root);
  return ans;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut i32) -> i32 {
    if root.is_none() {
      return 0;
    }
    let node = root.as_ref().unwrap().borrow();
    let left = (0).max(Self::dfs(&node.left, res));
    let right = (0).max(Self::dfs(&node.right, res));
    *res = (node.val + left + right).max(*res);
    node.val + left.max(right)
  }

  pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    let mut res = -1000;
    Self::dfs(&root, &mut res);
    res
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxPathSum = function (root) {
  let ans = -1001;
  const dfs = root => {
    if (!root) {
      return 0;
    }
    const left = Math.max(0, dfs(root.left));
    const right = Math.max(0, dfs(root.right));
    ans = Math.max(ans, left + right + root.val);
    return Math.max(left, right) + root.val;
  };
  dfs(root);
  return ans;
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
public class Solution {
  private int ans = -1001;

  public int MaxPathSum(TreeNode root) {
    dfs(root);
    return ans;
  }

  private int dfs(TreeNode root) {
    if (root == null) {
      return 0;
    }
    int left = Math.Max(0, dfs(root.left));
    int right = Math.Max(0, dfs(root.right));
    ans = Math.Max(ans, left + right + root.val);
    return root.val + Math.Max(left, right);
  }
}