742 Closest Leaf in a Binary Tree

742. Closest Leaf in a Binary Tree​
1. Question
Given a binary tree where every node has a unique value, and a target keyk, find the value of the nearest leaf node to targetkin the tree.
Here,nearestto a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called aleafif it has no children.
In the following examples, the input tree is represented in flattened form row by row. The actualroottree given will be a TreeNode object.
Example 1:
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Input: root = [1, 3, 2], k = 1
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Diagram of binary tree:
3
          1
4
         / \
5
        3   2
6
​
7
​
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Output: 2 (or 3)
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​
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Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.
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Example 2:
1
Input: root = [1], k = 1
2
​
3
Output: 1
4
​
5
​
6
Explanation: The nearest leaf node is the root node itself.
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Example 3:
1
Input: root = [1,2,3,4,null,null,null,5,null,6], k = 2
2
Diagram of binary tree:
3
             1
4
            / \
5
           2   3
6
          /
7
         4
8
        /
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       5
10
      /
11
     6
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​
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​
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Output: 3
15
​
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Explanation:
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The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.
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Note:
1.rootrepresents a binary tree with at least1node and at most1000nodes.
2.Every node has a uniquenode.valin range[1, 1000].
3.There exists some node in the given binary tree for whichnode.val == k.
2. Implementation
(1) DFS + BFS
思路: 1. DFS to construct graph
1.BFS to find the closet leaf node
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class Solution {
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    public int findClosestLeaf(TreeNode root, int k) {
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        TreeNode[] start = new TreeNode[1];
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        Map<TreeNode, List<TreeNode>> adjList = new HashMap<>();
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        buildGraph(root, null, k, start, adjList);
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​
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        Queue<TreeNode> queue = new LinkedList<>();
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        queue.add(start[0]);
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​
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        Set<TreeNode> visited = new HashSet<>();
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​
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        while (!queue.isEmpty()) {
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            int size = queue.size();
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​
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            for (int i = 0; i < size; i++) {
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                TreeNode curNode = queue.remove();
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                visited.add(curNode);
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​
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                if (curNode.left == null && curNode.right == null) {
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                    return curNode.val;
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                }
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​
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                for (TreeNode nextNode : adjList.get(curNode)) {
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                    if (!visited.contains(nextNode)) {
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                        queue.add(nextNode);
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                    }
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                }
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            }
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        }
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        return -1;
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​
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    }
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​
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    public void buildGraph(TreeNode curNode, TreeNode parent, int k, TreeNode[] start, Map<TreeNode, List<TreeNode>> adjList) {
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        if (curNode == null) {
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            return;
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        }
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​
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        if (curNode.val == k) {
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            start[0] = curNode;
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        }
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​
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        if (parent != null) {
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            adjList.putIfAbsent(curNode, new ArrayList<>());
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            adjList.putIfAbsent(parent, new ArrayList<>());
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            adjList.get(curNode).add(parent);
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            adjList.get(parent).add(curNode);
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        }
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​
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        buildGraph(curNode.left, curNode, k, start, adjList);
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        buildGraph(curNode.right, curNode, k, start, adjList);
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    }
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}
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3. Time & Space Complexity
DFS + BFS : 时间复杂度O(n), 空间复杂度O(n)

655 Print Binary Tree

783 Minimum Distance Between BST Nodes

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Contents

742. Closest Leaf in a Binary Tree

1. Question

2. Implementation

3. Time & Space Complexity

742. Closest Leaf in a Binary Tree​

1. Question

2. Implementation

3. Time & Space Complexity

742. Closest Leaf in a Binary Tree