Tree110BalancedBinaryTree

Attention

the definition of “balanced” here is for “the depths of the two sub trees”, not “the depths of all the leaves”.

思路

• Level Traversal, check the number of each level equals 2^height Or not. Failed Case 2
  
  • Actually it is ’ the depths of all leaves’ as Attention listed.
• Fine the node that ‘made’ the differ of height
  
  • If one of its child is null, check its child’s child exist or not.
  • Logic is wrong. See Fail Case 1
• Return the height of each child.
  
  • Check itself is balanced and check its child
  • Recursive one takes 3ms
  • Actually we can use non-recursive one. The only reason to use it is because —– The “raw content”干货 of this function is very small, just comparing the height of two sub-trees.

Failed case

1. 只考虑了difference在同一个节点的情况，没考虑不同节点。思路2不正确
   
Input: [1,2,2,3,3,null,null,4,4]
Output: true
Expected: false

2. 思路1也不正确，并没有真正读题，以下情况应该返回1
   
Input:[1,2,2,3,3,3,3,4,4,4,4,4,4,null,null,5,5]
Output:false
Expected:true

   

TIPS

• 一旦看到IfElse的return很简单，可以写成？：形式，这样比较简明，不然太繁杂
• 叠加的else可以简化

Original getDepth()

private static int getDepth(TreeNode root) {        if (root == null) return 0;        int leftDepth = getDepth(root.left);        //If the left child is balanced, we search the other one        if (leftDepth != -1) {            int rightDepth = getDepth(root.right);            //If the right child is balanced            if (rightDepth != -1) {                if (Math.abs(leftDepth - rightDepth) > 1 ) {return -1;}                else { return 1 + Math.max(leftDepth, rightDepth);}            } else {                return -1;//This can be simplized in the end of the function.            }        } else {            return -1;        }    }

简化后为

    private static int getDepth(TreeNode root) {        if (root == null) return 0;        int leftDepth = getDepth(root.left);        //If the left child is balanced, we search the other one        if (leftDepth != -1) {            int rightDepth = getDepth(root.right);            //If the right child is balanced            if (rightDepth != -1) {                return Math.abs(leftDepth - rightDepth) > 1 ?  -1 :  1 + Math.max(leftDepth, rightDepth);            }        }        return -1;    }