12
05/2015
[LeetCode] Balanced Binary Tree
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.
解题思路：
先检查左孩子树的高度与右孩子树的高度是否相差大于1。若大于1，则返回false，否则检查左孩子树和右孩子树是否均是平衡树。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode* root) {
if(root==NULL){
return true;
}
int leftH = TreeHeight(root->left);
int rightH = TreeHeight(root->right);
if(leftH > rightH + 1 || leftH < rightH -1){
return false;
}
return isBalanced(root->left) && isBalanced(root->right);
}
int TreeHeight(TreeNode* root){
if(root==NULL){
return 0;
}else{
return 1+max(TreeHeight(root->left), TreeHeight(root->right));
}
}
};二次刷题（2015-08-04）
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode* root) {
if(root==NULL){
return true;
}
int leftHeight = getHeight(root->left);
int rightHeight = getHeight(root->right);
if(abs(leftHeight - rightHeight)>1){
return false;
}
return isBalanced(root->left) && isBalanced(root->right);
}
int getHeight(TreeNode* root){
if(root==NULL){
return 0;
}
return 1 + max(getHeight(root->left), getHeight(root->right));
}
};转载请注明：康瑞部落 » [LeetCode] Balanced Binary Tree
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