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Binary Tree: Path Sum Iterative Post Order approach and explanation

Module: Binary Tree

You can refer to the Leetcode problem 112. Path Sum

Problem Statement

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22

Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Post Order and path sum



Proceed in a classic binary-tree-postorder-traversal iterative fashion, with following 2 things in mind:

  1. Keep a currSum variable which will have the total sum added up to curr node.
  2. On each pop-operation/climb-up (processing of a node), and since we know that we have already checked if this node was a leaf node and if the currSum till this node is equal to target, we reduce it's value from currSum which we added while traversing downwards.

While you descend and go on adding the node values, on traversing up, we need to discard added sum of those nodes.


Input: [1,-2,-3,1,3,-2,null,-1]
targetSum: 2


Steps/Dry run :

  1. We declared a currSum variable to hold our total sum till current node
  2. We kept a pre reference variable which will point to a previous node that has been last popped-out or last processed. This will help us not to re-process the right child.
  3. We start scanning from root node and till we reach a left-most node, for each node, do
    • Add curr.val to currSum
    • push the curr node onto stack s
    • traverse to left child
  4. When the inner while loop ends, our curr would be pointing to null (left-child of the left-most node, this left-most node would always be the leaf node).
  5. Our curr is null, there is nothing to go left now, start looking into our stack
  6. peek from stack i.e. curr = s.peek. Remember, we are doing a post-order traversal, so we need to descend right from this peeked node, and hence is the peek operation, we can not just pop from stack.
  7. The very first peek from stack would give us the left-most node(a leaf node). This is not important, but the point is at this point, we start climbing up.
  8. Check if this node meets our criteria of a leaf node with sum till this equals to target.
    • If yes, we are done with the whole process, no need to scan further and we can just break from this point.
  9. If the flow continues i.e. even if it was a leaf node but the sum till this node is not what we are looing for, start popping out.
  10. We only pop(post-order) if any of following 2 conditions are met
    • If there is no right-child
    • If we already processed the right child(right child is equal/point to pre)
  11. If the above condition is true, and pop the current node, do following:
    • Decrease the currSum as currSum -= curr.val. (We need to revert what we added to our currSum)
    • point pre to this curr node.
    • point curr to null
  12. If we have not processed the right child, descend to right child. (and the loop again begins with this right child as new curr and will look for the left most node of this curr and go on adding the values to currSum.)

So, in above example, once we are done verifying node g(it's a leaf node and currSum != targetSum), we shall be climbing up to node d then b and then to a leaf node e. While climbing up from g -> d -> b -> e, we descrease the corresponding values of node g and d. Note that we are not decreasing the value of b as we are not yet done with right child e.

In case our currSum would have been -2, then node e would not meet our criteria and we would be descreasing the values of node e and b also.

Time : O(n)
Space : O(n


private boolean hasPathSum_iterative(TreeNode root, int targetSum){

        boolean hasPath = false;

        if(root == null)
            return hasPath;

        TreeNode curr = root;
        TreeNode pre = null;
        int currSum = 0;
        Stack<TreeNode> s = new Stack();

        while(curr != null || !s.isEmpty()){

            while(curr != null){
                currSum += curr.val;
                curr = curr.left;

            curr = s.peek();

            if(curr.left == null && curr.right == null && currSum == targetSum){
                hasPath = true;

            // If right ST is null or already processed(equal to pre)
            // then process curr
            if(curr.right == null || curr.right == pre){
                curr = s.pop();
                currSum -= curr.val;
                pre = curr;
                curr = null;   
            }else {
                curr = curr.right;


        return hasPath;
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Hope this might have help you to understand the concept of iterative post order and apply extra logic to solve the path-sum problem.

Indeed, there are various other approaches to do the same, I wrote this to make someone relate idea behind post-order and sum.

See also binary-tree-postorder-traversal

Problem Credit :

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