110. Balanced Binary Tree

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.

方法一、递归的方法

（1）从根节点开始得遍历会导致很多节点的多次遍历，代码如下：

 int treeDepth(TreeNode* root)    {        if(root == NULL)            return 0;                queue<TreeNode*> que;        que.push(root);        int countLevNodes = 1;        int depth = 0;        while(!que.empty())        {            TreeNode* temp = que.front();            que.pop();            countLevNodes--;            if(temp->left != NULL)                que.push(temp->left);            if(temp->right != NULL)                que.push(temp->right);            if(countLevNodes == 0)            {                depth++;                countLevNodes = que.size();            }        }                return depth;    }        bool isBalanced(TreeNode* root) {        if(NULL == root)            return true;                int left = treeDepth(root->left);        int right = treeDepth(root->right);        int diff = abs(left - right);        if(diff > 1)            return false;        return isBalanced(root->left)&&isBalanced(root->right);    }

（2）从后序开始遍历，使得很多节点的遍历只用遍历一次

采用的后续遍历的方式遍历二叉树的每一个节点的，在遍历到一个节点之前我们就已经遍历了它的左右子树。只要在遍历每个节点的时候记录它的深度（某一节点的深度等于

它到叶结点的路径的长度），我们就可以一边遍历一边判断每个节点是不是平衡的

bool isBalanced(TreeNode* root,int* pdepth)    {        if(NULL == root)        {            *pdepth = 0;            return true;        }                int left,right;        if(isBalanced(root->left,&left)&&isBalanced(root->right,&right))        {            int diff = left - right;            if(abs(diff)<=1)            {                *pdepth = 1 + (left>right?left:right);                return true;            }        }        return false;    }        bool isBalanced(TreeNode* root) {        int depth = 0;        return isBalanced(root,&depth);//由于每一次调用该函数都要去改变depth的数值，所以depth为引用传递    }

方法二、采用dfs的方法来计算每个节点的左右孩子节点的高度

int dfsHeight (TreeNode *root) {        if(NULL == root)            return 0;        int leftHeight = dfsHeight(root->left);        if(leftHeight == -1) return -1;        int rightHeight = dfsHeight(root->right);        if(rightHeight == -1) return -1;        if (abs(leftHeight - rightHeight) > 1)  return -1;        return max (leftHeight, rightHeight) + 1;    }    bool isBalanced(TreeNode* root) {        return dfsHeight(root) != -1;    }