CPMagical CP Library

Problem solution · Java

Closest Leaf in a Binary Tree

Closest Leaf in a Binary Tree: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
55 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Closest Leaf in a Binary Tree, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 55 lines of Java from the credited upstream file 742.java.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeClosest Leaf in a Binary Tree · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int findClosestLeaf(TreeNode root, int k) {    ans = -1;    minDist = 1000;    // {node: distance to TreeNode(k)}    Map<TreeNode, Integer> nodeToDist = new HashMap<>();     getDists(root, k, nodeToDist);    getClosestLeaf(root, 0, nodeToDist);     return ans;  }   private int ans;  private int minDist;   private void getDists(TreeNode root, int k, Map<TreeNode, Integer> nodeToDist) {    if (root == null)      return;    if (root.val == k) {      nodeToDist.put(root, 0);      return;    }     getDists(root.left, k, nodeToDist);    if (nodeToDist.containsKey(root.left)) {      // The TreeNode(k) is in the left subtree.      nodeToDist.put(root, nodeToDist.get(root.left) + 1);      return;    }     getDists(root.right, k, nodeToDist);    if (nodeToDist.containsKey(root.right))      // The TreeNode(k) is in the right subtree.      nodeToDist.put(root, nodeToDist.get(root.right) + 1);  }   private void getClosestLeaf(TreeNode root, int dist, Map<TreeNode, Integer> nodeToDist) {    if (root == null)      return;    if (nodeToDist.containsKey(root))      dist = nodeToDist.get(root);    if (root.left == null && root.right == null) {      if (dist < minDist) {        minDist = dist;        ans = root.val;      }      return;    }     getClosestLeaf(root.left, dist + 1, nodeToDist);    getClosestLeaf(root.right, dist + 1, nodeToDist);  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗