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1740. Find Distance in a Binary Tree

Description

Given the root of a binary tree and two integers p and q, return the distance between the nodes of value p and value q in the tree.

The distance between two nodes is the number of edges on the path from one to the other.

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 0
Output: 3
Explanation: There are 3 edges between 5 and 0: 5-3-1-0.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 7
Output: 2
Explanation: There are 2 edges between 5 and 7: 5-2-7.

Example 3:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 5
Output: 0
Explanation: The distance between a node and itself is 0.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • 0 <= Node.val <= 109
  • All Node.val are unique.
  • p and q are values in the tree.

Solutions

Related: 236-Lowest-Common-Ancestor-of-a-Binary-Tree/

First, find the lowest common ancestor of the two nodes with values p and q. Next, calculate the distances from the lowest common ancestor to the two nodes with values p and q, respectively. Finally, calculate the sum of the two distances and return.

Alternative-1

hashmap to store distance of every node-pair during finding LCA, then just map look up.

Alternative-2

Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2 * Dist(root, lca)

  • ‘n1’ and ‘n2’ are the two given keys
  • ‘root’ is root of given Binary Tree.
  • ‘lca’ is lowest common ancestor of n1 and n2
  • Dist(n1, n2) is the distance between n1 and n2.
  • /**
 * 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 {
    public int findDistance(TreeNode root, int p, int q) {
        TreeNode g = lca(root, p, q);
        return dfs(g, p) + dfs(g, q);
    }

    private int dfs(TreeNode root, int v) {
        if (root == null) {
            return -1;
        }
        if (root.val == v) {
            return 0;
        }
        int left = dfs(root.left, v);
        int right = dfs(root.right, v);
        if (left == -1 && right == -1) {
            return -1;
        }
        return 1 + Math.max(left, right);
    }

    private TreeNode lca(TreeNode root, int p, int q) {
        if (root == null || root.val == p || root.val == q) {
            return root;
        }
        TreeNode left = lca(root.left, p, q);
        TreeNode right = lca(root.right, p, q);
        if (left == null) {
            return right;
        }
        if (right == null) {
            return left;
        }
        return root;
    }
}

  • /**
 * 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 findDistance(TreeNode* root, int p, int q) {
        TreeNode* g = lca(root, p, q);
        return dfs(g, p) + dfs(g, q);
    }

    TreeNode* lca(TreeNode* root, int p, int q) {
        if (!root || root->val == p || root->val == q) return root;
        TreeNode* left = lca(root->left, p, q);
        TreeNode* right = lca(root->right, p, q);
        if (!left) return right;
        if (!right) return left;
        return root;
    }

    int dfs(TreeNode* root, int v) {
        if (!root) return -1;
        if (root->val == v) return 0;
        int left = dfs(root->left, v);
        int right = dfs(root->right, v);
        if (left == -1 && right == -1) return -1;
        return 1 + max(left, right);
    }
};

  • # 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 findDistance(self, root: Optional[TreeNode], p: int, q: int) -> int:
        def lca(root, p, q):
            if root is None or root.val in [p, q]:
                return root
            left = lca(root.left, p, q)
            right = lca(root.right, p, q)
            if left is None:
                return right
            if right is None:
                return left
            return root

        def dfs(root, v):
            if root is None:
                return -1
            if root.val == v:
                return 0
            left, right = dfs(root.left, v), dfs(root.right, v)
            if left == right == -1:
                return -1
            return 1 + max(left, right)

        g = lca(root, p, q)
        return dfs(g, p) + dfs(g, q)


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findDistance(root *TreeNode, p int, q int) int {
	var lca func(root *TreeNode, p int, q int) *TreeNode
	lca = func(root *TreeNode, p int, q int) *TreeNode {
		if root == nil || root.Val == p || root.Val == q {
			return root
		}
		left, right := lca(root.Left, p, q), lca(root.Right, p, q)
		if left == nil {
			return right
		}
		if right == nil {
			return left
		}
		return root
	}
	var dfs func(root *TreeNode, v int) int
	dfs = func(root *TreeNode, v int) int {
		if root == nil {
			return -1
		}
		if root.Val == v {
			return 0
		}
		left, right := dfs(root.Left, v), dfs(root.Right, v)
		if left == -1 && right == -1 {
			return -1
		}
		return 1 + max(left, right)
	}
	g := lca(root, p, q)
	return dfs(g, p) + dfs(g, q)
}

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