leetcode之Maximum Depth of Binary Tree 以及Balanced Binary Tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

二叉树的深度，用递归特别方便！

代码如下：

public int maxDepth(TreeNode root) {
if(root==null){
return 0;
}
int le = maxDepth(root.left);
int ri = maxDepth(root.right);
return le > ri ? (le+1) : (ri+1);
}

　　扩展：判断一棵树是不是平衡二叉树？

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.

可以使用一般的方法，对每个结点求出左右深度，然后深度相减，绝对值是否小于1，但是这样会可能重复遍历一个节点多次。

代码一：虽然有重复计算，但是最终返回值那里很精妙。

　　　　public boolean isBalanced(TreeNode root) {

return maxDepth1(root) != -1;
}

private static int maxDepth1(TreeNode root) {
if (root == null) {
return 0;
}

int left = maxDepth1(root.left);
int right = maxDepth1(root.right);
if (left == -1 || right == -1 || Math.abs(left-right) > 1) {
return -1;
}
return Math.max(left, right) + 1;

}

　　代码2：利用了TreeNode结构中的val，用它来记录以当前结点为根的子树的高度，避免多次计算。

　　public boolean isBalanced(TreeNode root) {
height(root);
return run(root);
}

public boolean run(TreeNode root) {
if (root == null) return true;

int l = 0, r = 0;
if (root.left != null) l = root.left.val;
if (root.right != null) r = root.right.val;
if (Math.abs(l - r) <= 1 && isBalanced(root.left) && isBalanced(root.right)) return true;

return false;
}

public int height(TreeNode root) {
if (root == null) return 0;
root.val = Math.max( height(root.left), height(root.right) ) + 1;
return root.val;
}