Force buckle 110. Balanced binary tree

Portal 110. Balanced binary trees

Their thinking

Comparing the depth difference between the left and right subtrees of each node, we must recursively find the depth of the left and right subtrees , After finding the depth of both left and right nodes, directly compare , If false Just put flag Marked as false, If there is no mark at the end of the program , It shows that the binary tree is a balanced binary tree

Code

class Solution {
    
    boolean flag = true;
    public int getSize(TreeNode node){
    
        if(node == null)return 0;
        int lSize = getSize(node.left);
        int rSize = getSize(node.right);
        if(Math.abs(lSize - rSize) > 1)flag = false;
        return 1 + Math.max(lSize,rSize);
    }

    public boolean isBalanced(TreeNode root) {
    
        if(root == null)return true;
        getSize(root);
        return flag;
    }
}

• Time complexity O(n);
• Spatial complexity O(n);// Recursive space