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742. Closest Leaf in a Binary Tree πŸ”’

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

Solution 1: DFS + BFS

First, we use depth-first search to construct an undirected graph $g$, where $g[node]$ represents the set of nodes adjacent to the node $node$. Then we start a breadth-first search from node $k$ until we find a leaf node, which is the answer.

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

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# 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: Optional[TreeNode], k: int) -> int:
        def dfs(root: Optional[TreeNode], fa: Optional[TreeNode]):
            if root:
                g[root].append(fa)
                g[fa].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)
        vis = set(q)
        while 1:
            node = q.popleft()
            if node:
                if node.left == node.right:
                    return node.val
                for nxt in g[node]:
                    if nxt not in vis:
                        vis.add(nxt)
                        q.append(nxt)

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/**
 * 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 = new HashMap<>();

    public int findClosestLeaf(TreeNode root, int k) {
        dfs(root, null);
        Deque<TreeNode> q = new LinkedList<>();
        Set<TreeNode> vis = new HashSet<>(q.size());
        for (TreeNode node : g.keySet()) {
            if (node != null && node.val == k) {
                vis.add(node);
                q.offer(node);
                break;
            }
        }
        while (true) {
            TreeNode node = q.poll();
            if (node != null) {
                if (node.left == node.right) {
                    return node.val;
                }
                for (TreeNode nxt : g.get(node)) {
                    if (vis.add(nxt)) {
                        q.offer(nxt);
                    }
                }
            }
        }
    }

    private void dfs(TreeNode root, TreeNode fa) {
        if (root != null) {
            g.computeIfAbsent(root, k -> new ArrayList<>()).add(fa);
            g.computeIfAbsent(fa, k -> new ArrayList<>()).add(root);
            dfs(root.left, root);
            dfs(root.right, root);
        }
    }
}

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/**
 * 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 findClosestLeaf(TreeNode* root, int k) {
        unordered_map<TreeNode*, vector<TreeNode*>> g;
        function<void(TreeNode*, TreeNode*)> dfs = [&](TreeNode* root, TreeNode* fa) {
            if (root) {
                g[root].push_back(fa);
                g[fa].push_back(root);
                dfs(root->left, root);
                dfs(root->right, root);
            }
        };
        dfs(root, nullptr);
        queue<TreeNode*> q;
        unordered_set<TreeNode*> vis;
        for (auto& [node, _] : g) {
            if (node && node->val == k) {
                q.push(node);
                vis.insert(node);
            }
        }
        while (1) {
            auto node = q.front();
            q.pop();
            if (node) {
                if (node->left == node->right) {
                    return node->val;
                }
                for (auto& nxt : g[node]) {
                    if (vis.count(nxt)) {
                        continue;
                    }
                    q.push(nxt);
                    vis.insert(nxt);
                }
            }
        }
    }
};

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

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