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742. Closest Leaf in a Binary Tree

Description

Given the root of a binary tree where every node has a unique value and a target integer k, return the value of the nearest leaf node to the target k in the tree.

Nearest to a leaf means the least number of edges traveled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.

 

Example 1:

Input: root = [1,3,2], k = 1
Output: 2
Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.

Example 2:

Input: root = [1], k = 1
Output: 1
Explanation: The nearest leaf node is the root node itself.

Example 3:

Input: root = [1,2,3,4,null,null,null,5,null,6], k = 2
Output: 3
Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • 1 <= Node.val <= 1000
  • All the values of the tree are unique.
  • There exist some node in the tree where Node.val == k.

Solutions

DFS & BFS.

  • /**
 * 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 Map<TreeNode, List<TreeNode>> g;

    public int findClosestLeaf(TreeNode root, int k) {
        g = new HashMap<>();
        dfs(root, null);
        Deque<TreeNode> q = new LinkedList<>();
        for (Map.Entry<TreeNode, List<TreeNode>> entry : g.entrySet()) {
            if (entry.getKey() != null && entry.getKey().val == k) {
                q.offer(entry.getKey());
                break;
            }
        }
        Set<TreeNode> seen = new HashSet<>();
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            seen.add(node);
            if (node != null) {
                if (node.left == null && node.right == null) {
                    return node.val;
                }
                for (TreeNode next : g.get(node)) {
                    if (!seen.contains(next)) {
                        q.offer(next);
                    }
                }
            }
        }
        return 0;
    }

    private void dfs(TreeNode root, TreeNode p) {
        if (root != null) {
            g.computeIfAbsent(root, k -> new ArrayList<>()).add(p);
            g.computeIfAbsent(p, k -> new ArrayList<>()).add(root);
            dfs(root.left, root);
            dfs(root.right, 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:
    unordered_map<TreeNode*, vector<TreeNode*>> g;

    int findClosestLeaf(TreeNode* root, int k) {
        dfs(root, nullptr);
        queue<TreeNode*> q;
        for (auto& e : g) {
            if (e.first && e.first->val == k) {
                q.push(e.first);
                break;
            }
        }
        unordered_set<TreeNode*> seen;
        while (!q.empty()) {
            auto node = q.front();
            q.pop();
            seen.insert(node);
            if (node) {
                if (!node->left && !node->right) return node->val;
                for (auto next : g[node]) {
                    if (!seen.count(next))
                        q.push(next);
                }
            }
        }
        return 0;
    }

    void dfs(TreeNode* root, TreeNode* p) {
        if (!root) return;
        g[root].push_back(p);
        g[p].push_back(root);
        dfs(root->left, root);
        dfs(root->right, root);
    }
};

  • # 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 findClosestLeaf(self, root: TreeNode, k: int) -> int:
        def dfs(root, p):
            if root:
                g[root].append(p)
                g[p].append(root)
                dfs(root.left, root)
                dfs(root.right, root)

        g = defaultdict(list)
        dfs(root, None)
        q = deque([node for node in g if node and node.val == k])
        seen = set()
        while q:
            node = q.popleft()
            seen.add(node)
            if node:
                if node.left is None and node.right is None:
                    return node.val
                for next in g[node]:
                    if next not in seen:
                        q.append(next)


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findClosestLeaf(root *TreeNode, k int) int {
	g := make(map[*TreeNode][]*TreeNode)
	var dfs func(root, p *TreeNode)
	dfs = func(root, p *TreeNode) {
		if root == nil {
			return
		}
		g[root] = append(g[root], p)
		g[p] = append(g[p], root)
		dfs(root.Left, root)
		dfs(root.Right, root)
	}
	dfs(root, nil)
	var q []*TreeNode
	for t, _ := range g {
		if t != nil && t.Val == k {
			q = append(q, t)
			break
		}
	}
	seen := make(map[*TreeNode]bool)
	for len(q) > 0 {
		node := q[0]
		q = q[1:]
		seen[node] = true
		if node != nil {
			if node.Left == nil && node.Right == nil {
				return node.Val
			}
			for _, next := range g[node] {
				if !seen[next] {
					q = append(q, next)
				}
			}
		}
	}
	return 0
}

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