LeetCode之Balanced Binary Tree
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/*遍历整颗二叉树,对于每遍历到的一点,都求它左右子树的深度。如果左右 子树的深度差小于或等于1,则平衡的。判断二叉树所有节点,如果有一个不是平衡的, 则不是平衡二叉树,否则为平衡二叉树。*/class Solution {public: bool isBalanced(TreeNode* root) { if(root == nullptr) return true; if(abs(depth(root->left) - depth(root->right)) > 1) return false; return isBalanced(root->left) && isBalanced(root->right); } int depth(TreeNode *root){ if(root == nullptr) return 0; int left(0), right(0); if(root->left != nullptr) left = depth(root->left); if(root->right != nullptr) right = depth(root->right); return max(left, right) + 1; }};
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