Solve Tree Problems Recursively
"Top-down" Solution
Here is the pseudocode for the recursion function maximum_depth(root, depth)
:
1. return if root is null 2. if root is a leaf node: 3. answer = max(answer, depth) // update the answer if needed 4. maximum_depth(root.left, depth + 1) // call the function recursively for left child 5. maximum_depth(root.right, depth + 1) // call the function recursively for right child
code:
int answer; // don't forget to initialize answer before call maximum_depth void maximum_depth(TreeNode* root, int depth) { if (!root) { return; } if (!root->left && !root->right) { answer = max(answer, depth); } maximum_depth(root->left, depth + 1); maximum_depth(root->right, depth + 1); }
"Bottom-up" Solution
1. return 0 if root is null // return 0 for null node 2. left_depth = maximum_depth(root.left) 3. right_depth = maximum_depth(root.right) 4. return max(left_depth, right_depth) + 1 // return depth of the subtree rooted at root
code:
int maximum_depth(TreeNode* root) { if (!root) { return 0; // return 0 for null node } int left_depth = maximum_depth(root->left); int right_depth = maximum_depth(root->right); return max(left_depth, right_depth) + 1; // return depth of the subtree rooted at root }
Conclusion.
It is not easy to understand recursion and find out a recursion solution for the problem.
When you meet a tree problem, ask yourself two questions: can you determine some parameters to help the node know the answer of itself? Can you use these parameters and the value of the node itself to determine what should be the parameters parsing to its children? If the answers are both yes, try to solve this problem using a "top-down
" recursion solution.
Or you can think the problem in this way: for a node in a tree, if you know the answer of its children, can you calculate the answer of the node? If the answer is yes, solving the problem recursively from bottom up
might be a good way.
In the following sections, we provide several classic problems for you to help you understand tree structure and recursion better.