Question

1026. Maximum Difference Between Node and Ancestor

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Description

Given the root of a binary tree, find the maximum value v for which there exist different nodes a and b where v = |a.val - b.val| and a is an ancestor of b.

A node a is an ancestor of b if either: any child of a is equal to b or any child of a is an ancestor of b.

 

Example 1:

Input: root = [8,3,10,1,6,null,14,null,null,4,7,13]
Output: 7
Explanation: We have various ancestor-node differences, some of which are given below :
|8 - 3| = 5
|3 - 7| = 4
|8 - 1| = 7
|10 - 13| = 3
Among all possible differences, the maximum value of 7 is obtained by |8 - 1| = 7.

Example 2:

Input: root = [1,null,2,null,0,3]
Output: 3

 

Constraints:

• The number of nodes in the tree is in the range [2, 5000].
• 0 <= Node.val <= 105

Solutions

Solution 1

# 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 maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
    def dfs(root, mi, mx):
      if root is None:
        return
      nonlocal ans
      ans = max(ans, abs(mi - root.val), abs(mx - root.val))
      mi = min(mi, root.val)
      mx = max(mx, root.val)
      dfs(root.left, mi, mx)
      dfs(root.right, mi, mx)

    ans = 0
    dfs(root, root.val, root.val)
    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;

  public int maxAncestorDiff(TreeNode root) {
    dfs(root, root.val, root.val);
    return ans;
  }

  private void dfs(TreeNode root, int mi, int mx) {
    if (root == null) {
      return;
    }
    int x = Math.max(Math.abs(mi - root.val), Math.abs(mx - root.val));
    ans = Math.max(ans, x);
    mi = Math.min(mi, root.val);
    mx = Math.max(mx, root.val);
    dfs(root.left, mi, mx);
    dfs(root.right, mi, mx);
  }
}

/**
 * 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 maxAncestorDiff(TreeNode* root) {
    int ans = 0;
    function<void(TreeNode*, int, int)> dfs = [&](TreeNode* root, int mi, int mx) {
      if (!root) {
        return;
      }
      ans = max({ans, abs(mi - root->val), abs(mx - root->val)});
      mi = min(mi, root->val);
      mx = max(mx, root->val);
      dfs(root->left, mi, mx);
      dfs(root->right, mi, mx);
    };
    dfs(root, root->val, root->val);
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func maxAncestorDiff(root *TreeNode) (ans int) {
  var dfs func(*TreeNode, int, int)
  dfs = func(root *TreeNode, mi, mx int) {
    if root == nil {
      return
    }
    ans = max(ans, max(abs(mi-root.Val), abs(mx-root.Val)))
    mi = min(mi, root.Val)
    mx = max(mx, root.Val)
    dfs(root.Left, mi, mx)
    dfs(root.Right, mi, mx)
  }
  dfs(root, root.Val, root.Val)
  return
}

func abs(x int) int {
  if x < 0 {
    return -x
  }
  return x
}

/**
 * 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 maxAncestorDiff(root: TreeNode | null): number {
  const dfs = (root: TreeNode | null, mi: number, mx: number): void => {
    if (!root) {
      return;
    }
    ans = Math.max(ans, Math.abs(root.val - mi), Math.abs(root.val - mx));
    mi = Math.min(mi, root.val);
    mx = Math.max(mx, root.val);
    dfs(root.left, mi, mx);
    dfs(root.right, mi, mx);
  };
  let ans: number = 0;
  dfs(root, root.val, root.val);
  return ans;
}

/**
 * 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 maxAncestorDiff = function (root) {
  let ans = 0;
  const dfs = (root, mi, mx) => {
    if (!root) {
      return;
    }
    ans = Math.max(ans, Math.abs(mi - root.val), Math.abs(mx - root.val));
    mi = Math.min(mi, root.val);
    mx = Math.max(mx, root.val);
    dfs(root.left, mi, mx);
    dfs(root.right, mi, mx);
  };
  dfs(root, root.val, root.val);
  return ans;
};