【算法题】判断二叉树平衡性

由平衡二叉树的定义可知，每个节点的左右子树的高度差要小于等于1

• 可以用递归实现，在遍历二叉树各节点事，若节点的左右子树的深度之差不超过1，则是平衡二叉树。

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int getDepth(Tree*root){    if (root == NULL)        return 0;    int leftDepth = getDepth(root->left);    int rightDepth = getDepth(root->right);    return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;}bool IsBalanced(Tree* root){    if (root == NULL)        return true;    int leftDepth = getDepth(root->left);    int rightDepth = getDepth(root->right);    int diff = leftDepth - rightDepth;    if (abs(diff) > 1)        return false;    return IsBalanced(root->left) && IsBalanced(root->right);}

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• 上述方法在求该结点的的左右子树深度时遍历一遍树，再次判断子树的平衡性时又遍历一遍树结构，造成遍历多次。因此方法二是一边遍历树一边判断每个结点是否具有平衡性

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bool IsBalancedCore(Tree* root, int& depth){    if (root == NULL)    {        depth = 0;        return true;    }    int leftdepth(0), rightdepth(0);    bool leftbalanced = IsBalancedCore(root->left, leftdepth);    bool rightbalanced = IsBalancedCore(root->right, rightdepth);    bool rootbalenced = abs(leftbalanced - rightbalanced) < 2;    depth = max(leftdepth, rightdepth)+1;    return leftbalanced&&rightbalanced&&rootbalenced;}bool IsBalanced(Tree* root){    int i(0);    return IsBalancedCore(root, i);}

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