劍指offer 面試55題

面試55題：

題目：二叉樹的深度

題：輸入一棵二叉樹，求該樹的深度。從根結點到葉結點依次經過的結點（含根、葉結點）形成樹的一條路徑，最長路徑的長度為樹的深度。

 

解題思路：

①如果一棵樹只有一個節點，它的深度為1

②如果根節點只有左子樹而沒有右子樹，那么樹的深度是左子樹的深度加1

同樣，如果根節點只有右子樹而沒有左子樹，那么樹的深度是右子樹的深度加1

既有右子樹又有左子樹時，數的深度是左子樹和右子樹深度較大者加1

利用遞歸很容易實現上述思路：

 

解題代碼：

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def TreeDepth(self, pRoot):
        # write code here
        if pRoot is None:
            return 0
        left = self.TreeDepth(pRoot.left)
        right = self.TreeDepth(pRoot.right)
        return max(left,right)+1

 

題目拓展：平衡二叉樹

輸入一棵二叉樹，判斷該二叉樹是否是平衡二叉樹。

 

解題代碼：

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:  
    def TreeDepth(self, pRoot):
        if pRoot is None:
            return 0
        left = self.TreeDepth(pRoot.left)
        right = self.TreeDepth(pRoot.right)
        return max(left,right)+1
    
    def IsBalanced_Solution(self, pRoot):
        # write code here
        if pRoot is None:
            return True
        left = self.TreeDepth(pRoot.left)
        right = self.TreeDepth(pRoot.right)
        diff = left - right
        if diff< -1 or diff >1:
            return False
        return self.IsBalanced_Solution(pRoot.left) and self.IsBalanced_Solution(pRoot.right)