LeetCode – Minimum Depth of Binary Tree (Java)

Category: Algorithms February 5, 2013

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Thoughts

LinkedList is a queue in Java. The add() and remove() methods are used to manipulate the queue.

Java Solution

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int minDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
 
        LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
        LinkedList<Integer> counts = new LinkedList<Integer>();
 
        nodes.add(root);
        counts.add(1);
 
        while(!nodes.isEmpty()){
            TreeNode curr = nodes.remove();
            int count = counts.remove();
 
            if(curr.left == null && curr.right == null){
                return count;
            }
 
            if(curr.left != null){
                nodes.add(curr.left);
                counts.add(count+1);
            }
 
            if(curr.right != null){
                nodes.add(curr.right);
                counts.add(count+1);
            }
        }
 
        return 0;
    }
}

Category >> Algorithms

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Alik Elzin

2 different solutions:
int minDepthIterative(Node head) { if (n == null) return 0;
int depth = 1; Queue currLevel = new LinkedList(); currLevel.add(head); Queue nextLevel = new LinkedList(); while(true) { while(!currLevel.isEmpty()) { Node n = currLevel.remove(); if (n.left == null || n.right == null) return depth; nextLevel.add(n.left); nextLevel.add(n.right); } depth++; currLevel = nextLevel; nextLevel = new LinkedList(); } } /* This solution is problematic because it goes down all of the left tree where the right can be much shorter.*/ int minDepthRecurse(Node n) { if (n == null) return 0; if (n.left == null || n.right == null) return 1;
return Math.min(minDepth(n.left), minDepth(n.right)) + 1; }
Kishore Srinivas

Actually,this problem is not the exact reverse of the max depth where we select the maximum depth in each subtree + 1, your code will fail for the input [1,2] where 1 is root and 2 is the left child, it will return 1, but the answer is 2.
Haohua Wang

possible solution:

public static int minDepth(TreeNode root) { r
if (root == null) { return 0; }
return 1 + Math.min(minDepth(root.left), minDepth(root.right));
}
Manu Sinha

public int minDepth(TreeNode root) {
if(root == null){ return 0; } return helper(root, 1); } private int helper(TreeNode node, int level){ if(node.left == null && node.right == null){ return level; } int lDepth = Integer.MAX_VALUE; if(node.left != null){ lDepth = helper(node.left , level + 1); } int rDepth = Integer.MAX_VALUE; if(node.right != null){ rDepth = helper(node.right , level + 1); }
return Math.min(lDepth , rDepth); }
Iram22

static int min=Integer.MAX_VALUE; private static void minDepth(Node n,int x){ if(n==null) return; if(n.left==null && n.right==null){ min=Math.min(min,x+1); } minDepth(n.left,x+1); minDepth(n.right,x+1); }
Ankit Shah

shouldn’t it be || instead of &&
f(curr.left == null || curr.right == null){ return count;}
rupalph

Correct recursive solution:
public int minDepth(TreeNode node,int depth) {
if (node == null) return 0; int depth1=minDepth(node.left,depth+1); int depth2=minDepth(node.right,depth+1); return (Math.min(depth1,depth2 )+depth); }
Call above method as minDepth(root,0)
Israel Lopez

It works with only one queue

private int minimunDepth(TreeNode root){

int totalDepth =0;

if(root == null)
return totalDepth;

Queue queue = new LinkedList();
queue.add(root);
totalDepth++;
TreeNode rootNew, rootNewRight;

while (!queue.isEmpty()){

TreeNode node = queue.poll();
root = node;
totalDepth++;
rootNew = node.leftChild;
rootNewRight = node.rightChild;

if (rootNew != null && (rootNew.leftChild !=null || rootNew.rightChild!=null) )
queue.add(rootNew);
else break;
if (rootNewRight != null && (rootNewRight.rightChild != null || rootNewRight.leftChild != null) )
queue.add(rootNewRight);
else break;

}

return totalDepth;

}
juno

return Math.max(minDepth(root.left), minDepth(root.right)) +1;
ryanlr

Changed.
gao can

Seems wrong
Ankit Shah

you can further optimzer your solution Vishal by doing this: not need for the other code in between.

private int minDepth(TreeNode root) {
if (root == null) return 0;
return 1 + Math.min(minDepth(root.left), minDepth(root.right));
}
Yunxiao Zou

A simpler Java Solution:

public class Solution {

int minHeight = -1;

public int minDepth(TreeNode root) {

if (root == null)

return 0;

iterateTree(root, 1);

return minHeight;

}

private void iterateTree(TreeNode root, int i) {

if (root.left == null && root.right == null) {

if (minHeight == -1)

minHeight = i;

else if (minHeight > i)

minHeight = i;

}

if (root.left != null)

iterateTree(root.left, i + 1);

if (root.right != null)

iterateTree(root.right, i + 1);

}

}
Karthik

amazing solution ji
Vishal

public int minDepth(TreeNode root) {
if(root == null) return 0;
if(root.left ==null) return 1+minDepth(root.right);
if(root.right ==null) return 1+minDepth(root.left);
return 1+Math.min(minDepth(root.left),minDepth(root.right));
}
CopperCash

python version recursive min depth:

def min_depth(self, depth=1):
left, right = self.left, self.right
left_min = left.min_depth(depth + 1) if left else depth
right_min = right.min_depth(depth + 1) if right else depth
return min(left_min, right_min)
Windforce

I don’t understand the
return 0;
line. It seems the while loop should return the right answer. But as I tried, I can’t pass oj without this line. However, I changed it to ” return 1″ or “return 1000”. They all get passed. Weird.
Zhaoyu Li

I figured out a recursive solution, see my github: https://github.com/zlmoment/Leetcode-Solutions/blob/28d39da3002eb9f3ecf35d32cf87a145c0b58453/Minimum%20Depth%20of%20Binary%20Tree/MinimumDepthOfBinaryTree.java
Guest

Is this solution for Depth of tree or height of tree???
junmin

move thisif(curr.left == null && curr.right == null){ return count;} to the front of the other two ifs will optimize the solution

LeetCode – Minimum Depth of Binary Tree (Java)

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