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  • 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.
     
    复制代码
    /**
     * 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;
        }
    }
    复制代码

    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. 

    复制代码
    /**
     * 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);
        }
    }
    复制代码

    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.

    复制代码
    复制代码
    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);
        }
    } 
    复制代码
    复制代码
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  • 原文地址:https://www.cnblogs.com/timoBlog/p/4635193.html
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