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1026. Maximum Difference Between Node and Ancestor

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

  • /**
 * 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.
# 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.
 * 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;
};


  • /**
 * 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;

    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);
    }
}

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