# An iterative solution

• ``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
private void pushLeft(Stack<TreeNode> stack, TreeNode node, Stack<int[]> diameters) {
while (node != null) {
stack.push(node);
diameters.push(new int[]{1, 1});
node = node.left;
}
}

public int diameterOfBinaryTree(TreeNode root) {
if (root == null) {
return 0;
}
int max = Integer.MIN_VALUE;
Stack<TreeNode> stack = new Stack<>();
// <diameter with left and right, max diameter with left or right>
Stack<int[]> diameters = new Stack<>();
pushLeft(stack, root, diameters);
TreeNode pre = null;
int[] preRes = new int[2];
while (!stack.isEmpty()) {
TreeNode node = stack.peek();
int[] diameter = diameters.peek();
diameter[0] = diameter[0] + preRes[1];
diameter[1] = Math.max(diameter[1], 1 + preRes[1]);
if (node.right != null && node.right != pre) {
pushLeft(stack, node.right, diameters);
preRes = new int[2];
continue;
}
stack.pop();
diameters.pop();
max = Math.max(max, diameter[0]);
pre = node;
preRes = diameter;
}

return max - 1;
}
}
``````

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