LeetCode: Balanced Binary Tree 解题报告

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.

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SOLUTION 1:

使用inner Class来解决

 // Solution 1:
     public boolean isBalanced1(TreeNode root) {
         return dfs(root).isBalanced;
     }     // bug 1: inner class is like:  "public class ReturnType {", no ()
     public class ReturnType {
         boolean isBalanced;
         int depth;         ReturnType(int depth, boolean isBalanced) {
             this.depth = depth;
             this.isBalanced = isBalanced;
         }
     }     public ReturnType dfs(TreeNode root) {
         ReturnType ret = new ReturnType(0, true);         if (root == null) {
             return ret;
         }         ReturnType left = dfs(root.left);
         ReturnType right = dfs(root.right);         ret.isBalanced = left.isBalanced
                          && right.isBalanced
                          && Math.abs(left.depth - right.depth) <= 1;         // bug 2: remember to add 1( the root depth )
         ret.depth = Math.max(left.depth, right.depth) + 1;         return ret;
     }

SOLUTION 2:

将 get depth函数提出

 // Solution 2:
     public boolean isBalanced(TreeNode root) {
         if (root == null) {
             return true;
         }                 return isBalanced(root.left) && isBalanced(root.right)
             && Math.abs(getDepth(root.left) - getDepth(root.right)) <= 1;
     }     public int getDepth(TreeNode root) {
         if (root == null) {
             return 0;
         }         return Math.max(getDepth(root.left), getDepth(root.right)) + 1;
     }

SOLUTION 3:

leetcode又加强了数据，solution 2对于一条单链过不了了。所以主页君加了一点优化，当检测到某个子节点为null时，求另一个子树的depth时，及时退出，这

样就不会产生getdepth太深的问题：

 // Solution 2:
     public boolean isBalanced(TreeNode root) {
         if (root == null) {
             return true;
         }         boolean cut = false;
         if (root.right == null || root.left == null) {
             cut = true;
         }         return isBalanced(root.left) && isBalanced(root.right)
             && Math.abs(getDepth(root.left, cut) - getDepth(root.right, cut)) <= ;
     }     public int getDepth(TreeNode root, boolean cut) {
         if (root == null) {
             return -;
         }         if (cut && (root.left != null || root.right != null)) {
             // if another tree is not deep, just cut and return fast.
             // Improve the performance.
             return ;
         }         return  + Math.max(getDepth(root.left, false), getDepth(root.right, false));
     }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsBalanced.java