112_pathSum.md

112. Path Sum 🌟

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.

O(N) Time , recursive

• if root is null return false
• if roots left and right both are null return root->val==targetSum.
• else recursively find targetSum - root->val in left and right subtree.

class Solution {
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root==NULL)
            return false;
        else if(root->left==NULL && root->right==NULL)
            return root->val==targetSum;
        else
            return hasPathSum(root->left,targetSum-root->val)
            || hasPathSum(root->right,targetSum-root->val);
    }
};

O(N) Time , iterative

soon...

class Solution{
public:
    bool hasPathSum(TreeNode *root, int sum){
        stack<TreeNode *> st;
        TreeNode *pre = NULL, *curr = root;
        int pathSum = 0;

        while (curr || !st.empty()){
            while (curr){
                st.push(curr);
                pathSum += curr->val;
                curr = curr->left;
            }

            curr = st.top();
            if (curr->left == NULL && curr->right == NULL && pathSum == sum)
                return true;

            if (curr->right && pre != curr->right){
                curr = curr->right;
            }
            else{
                pre = curr;
                st.pop();
                pathSum -= curr->val;
                curr = NULL;
            }
        }
        return false;
    }
};