LeetCode: 104. Maximum Depth of Binary Tree-简单-Python+Java

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.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

3

/ \

9 20

/ \

15 7

return its depth = 3.

 

解题思路：

Recursion是最直接的方式：

如果这个node是null，那么就直接return 0

如果不是null，recursion 它的left node 和right node， 取两个值中的max值+1 就是最大深度

也可以用Iterative去解，一层一层去遍历元素，每遍历一层，高度加一，直到该层的元素为空

 

Java:

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

Python:

Recursion方式

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None:
            return 0
        else:
            return max(self.maxDepth(root.left),self.maxDepth(root.right))+1

Iterative方式

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        height = 0
        if root==None:
            return height;
        level = [root]
        while level:
            temp = []
            for val in level:
                if val.right:
                    temp.append(val.right)
                if val.left:
                    temp.append(val.left)
            level = temp
            height += 1
        return height