Diameter of Binary Tree
iven a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
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Example:
Given a binary tree
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/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
Java Solution
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int diameterOfBinaryTree(TreeNode root) {
if(root == null){
return 0;
}
int maxLeft = diameterOfBinaryTree(root.left);
int maxRight = diameterOfBinaryTree(root.right);
int max = maxLeft > maxRight ? maxLeft : maxRight;
int diameter = getDepth(root.left) + getDepth(root.right);
if(max < diameter){
return diameter;
}
else{
return max;
}
}
private int getDepth(TreeNode root){
if(root != null){
return Math.max(getDepth(root.left), getDepth(root.right)) + 1;
}
return 0;
}
}
Approach #1: Depth-First Search [Accepted] Intuition
Any path can be written as two arrows (in different directions) from some node, where an arrow is a path that starts at some node and only travels down to child nodes.
If we knew the maximum length arrows L, R for each child, then the best path touches L + R + 1 nodes.
Algorithm
Let’s calculate the depth of a node in the usual way: max(depth of node.left, depth of node.right) + 1. While we do, a path “through” this node uses 1 + (depth of node.left) + (depth of node.right) nodes. Let’s search each node and remember the highest number of nodes used in some path. The desired length is 1 minus this number.
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class Solution {
int ans;
public int diameterOfBinaryTree(TreeNode root) {
ans = 1;
depth(root);
return ans - 1;
}
public int depth(TreeNode node) {
if (node == null) return 0;
int L = depth(node.left);
int R = depth(node.right);
ans = Math.max(ans, L+R+1);
return Math.max(L, R) + 1;
}
}
Complexity Analysis
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Time Complexity: O(N)O(N). We visit every node once.
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Space Complexity: O(N)O(N), the size of our implicit call stack during our depth-first search.
