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

For this problem, a height-balanced binary tree is defined as:

Example 1:

Given the following tree 

[3,9,20,null,null,15,7]

3
/ \
9  20
/  \
15   7

Return true.

Example 2:

Given the following tree 

[1,2,2,3,3,null,null,4,4]

1
/ \
2   2
/ \
3   3
/ \
4   4

Return false.

给定一个二叉树，判断是否高度平衡。高度平衡二叉树的定义：二叉树的任意节点的两个子树的深度差不超过1。

解法：根据定义，只需要判定一颗二叉树的左右子树高度的高度差是否小于等于1。递归处理每一颗二叉树左右子树的高度，并进行判断再回溯。

Java:

class Solution {
int abs(int x) {
return x > 0 ? x : -x;
}

int check(TreeNode* root) {
if (!root) return NULL;

int lch = check(root -> left);
int rch = check(root -> right);
// 检查子树是否存在不平衡
if (lch == -1 || rch == -1 || abs(lch - rch) > 1) return -1;

// 返回当前子树高度
return (lch > rch ? lch : rch) + 1;
}
public:
bool isBalanced(TreeNode* root) {
return check(root) != -1;
}
};　　

Java:  without ResultType

public class Solution {
public boolean isBalanced(TreeNode root) {
return maxDepth(root) != -1;
}

private int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}

int left = maxDepth(root.left);
int right = maxDepth(root.right);
if (left == -1 || right == -1 || Math.abs(left-right) > 1) {
return -1;
}
return Math.max(left, right) + 1;
}
}　

Python:

class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution:
# @param root, a tree node
# @return a boolean
def isBalanced(self, root):
return (self.getHeight(root) >= 0)

def getHeight(self, root):
if root is None:
return 0
left_height, right_height = self.getHeight(root.left), self.getHeight(root.right)
if left_height < 0 or right_height < 0 or abs(left_height - right_height) > 1:
return -1
return max(left_height, right_height) + 1　

Python:

class Solution:
"""
@param root: The root of binary tree.
@return: True if this Binary tree is Balanced, or false.
"""
def isBalanced(self, root):
balanced, _ = self.validate(root)
return balanced

def validate(self, root):
if root is None:
return True, 0

balanced, leftHeight = self.validate(root.left)
if not balanced:
return False, 0
balanced, rightHeight = self.validate(root.right)
if not balanced:
return False, 0

return abs(leftHeight - rightHeight) <= 1, max(leftHeight, rightHeight) + 1　　

C++:

class Solution {
public:
bool isBalanced(TreeNode *root) {
if (!root) return true;
if (abs(getDepth(root->left) - getDepth(root->right)) > 1) return false;
return isBalanced(root->left) && isBalanced(root->right);
}
int getDepth(TreeNode *root) {
if (!root) return 0;
return 1 + max(getDepth(root->left), getDepth(root->right));
}
};

C++:

class Solution {
public:
bool isBalanced(TreeNode *root) {
if (checkDepth(root) == -1) return false;
else return true;
}
int checkDepth(TreeNode *root) {
if (!root) return 0;
int left = checkDepth(root->left);
if (left == -1) return -1;
int right = checkDepth(root->right);
if (right == -1) return -1;
int diff = abs(left - right);
if (diff > 1) return -1;
else return 1 + max(left, right);
}
};

C++:

/**
* Definition of TreeNode:
* class TreeNode {
* public:
*     int val;
*     TreeNode *left, *right;
*     TreeNode(int val) {
*         this->val = val;
*         this->left = this->right = NULL;
*     }
* }
*/
class Solution {
public:
int depth(TreeNode *root) {
if (root == NULL) {
return 0;
}
int left = depth(root->left);
int right = depth(root->right);
if (left == -1 || right == -1 || abs(left - right) > 1) {
return -1;
}
return max(left, right) + 1;
}

/**
* @param root: The root of binary tree.
* @return: True if this Binary tree is Balanced, or false.
*/
bool isBalanced(TreeNode *root) {
return depth(root) != -1;
}
};

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