Leetcode & CTCI ---Day 1

Maximum Depth of Binary Tree  (DFS, TREE)

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

I use simplest dfs algorithm for this. The code just find each node‘s deepest height.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null)
            return 0;
        int leftHeight = maxDepth(root.left);
        int rightHeight = maxDepth(root.right);
        return Math.max(leftHeight, rightHeight) + 1;
    }
}

Same Tree

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

Firstly, I use a more brute force method. I did not think a better way to deal with the one of the p, q is null issue, so I write a lot of if else.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null)
            return true;
        else if (p == null && q != null)
            return false;
        else if (p != null && q == null)
            return false;
        else if (p.val != q.val)
            return false;
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}

Later I find a better way to deal with this if else condition for p, q is null issue to make the code shorter and more clear.

public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null)
            return true;
        else if (p==null || q==null)
            return false;
        else if (p.val != q.val)
            return false;
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}