# Accepted - Iterative Java Solution

• Iterative solution. Basically keeping track of the node and its height using a map and then making decisions based on that.

``````public boolean isBalanced(TreeNode root) {
if(root == null){
return true;
}
Stack<TreeNode> nodeStack = new Stack<>();
// This keeps a track of the height calculated for each node.
// The height of a node in a tree is the length of a longest path from the node to a leaf.
// If it has two paths of leafs with different heights. The max of them is the height of the node in this tree.
Map<TreeNode,Integer> nodeHeightMap= new HashMap<>();
nodeStack.push(root);
while (!nodeStack.isEmpty()){

TreeNode node = nodeStack.peek();
// The reason we are checking the map is we are looking if the height of a node has been calculated previously.
if(node.left!=null && !nodeHeightMap.containsKey(node.left)){
nodeStack.push(node.left);
}
else if( node.right!=null && !nodeHeightMap.containsKey(node.right)){
nodeStack.push(node.right);
}
else {
TreeNode nodePopped = nodeStack.pop();
int leftSubTreeHeight = (nodePopped.left == null)?0:nodeHeightMap.get(nodePopped.left);
int rightSubTreeHeight = (nodePopped.right == null)?0:nodeHeightMap.get(nodePopped.right);
// Absolute value of the difference between the left and right subtree heights.
int resultantHeightofNode = Math.abs(leftSubTreeHeight-rightSubTreeHeight);
// If the left & right subtree heights differ by more than 1 then it's not balanced binary tree.
if(resultantHeightofNode > 1){
return false;
}
nodeHeightMap.put(nodePopped,Math.max(leftSubTreeHeight,rightSubTreeHeight)+1);
}
}
return true;
}
``````

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