110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

110.平衡二叉樹

給定一棵二叉樹，判定其是否是高度平衡的

在這個問題中，一棵高度平衡的二叉樹定義為兩棵子樹的高度差不超過1，並且兩棵子樹也分別是高度平衡的。

思路：很明顯的呼叫遞迴解決的題。分別遞迴判定左子樹和右子樹是否為高度平衡的，不是直接返回false，如果左右子樹都分別是高度平衡的，再判斷當前是否為高度平衡的。先遞迴獲取左子樹和右子樹的高度，然後根據高度差得到是否為高度平衡。需要注意的是遞迴終點是根為NULL時，直接返回true即可，這也與輸入的特殊情況一致。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 
 int getDepth(struct TreeNode* root){
     if (root == NULL) return 0;
     int depthl = getDepth(root->left);
     int depthr = getDepth(root->right);
     return depthl>depthr?depthl+1:depthr+1;
 }
 
bool isBalanced(struct TreeNode* root) {
    if (root == NULL) return true;
    if (!isBalanced(root->left)) return false;
    if (!isBalanced(root->right)) return false;
    int depthl = getDepth(root->left);
    int depthr = getDepth(root->right);
    int diff = depthl - depthr;
    if (diff < 2 && diff > -2){
        return true;
    }
    else return false;
}
 

# 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 isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root: return True
        if not self.isBalanced(root.left): return False
        if not self.isBalanced(root.right): return False
        depthl = self.getDepth(root.left)
        depthr = self.getDepth(root.right)
        diff = depthl - depthr
        if (diff < 2 and diff > -2): return True
        else: return False
        
    def getDepth(self, root):
        if not root: return 0
        depthl = self.getDepth(root.left)
        depthr = self.getDepth(root.right)
        return depthl+1 if depthl>depthr else depthr+1