Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.
Return the merged tree.
Note: The merging process must start from the root nodes of both trees.
Constraints
The number of nodes in both trees is in the range [0, 2000].
// Definition for a binary tree node.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) {
this.val = val;
}
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
/**
* Time complexity : O(m). A total of m nodes need to be traversed.
* Here, m represents the minimum number of nodes from the two given trees.
* Space complexity : O(m). The depth of the recursion tree can go upto m
* in the case of a skewed tree. In average case, depth will be O(logm).
*/
class Solution {
public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
if(root1 == null && root2 == null) {
return null;
}
if(root1 == null) {
return root2;
}
if(root2 == null) {
return root1;
}
root1.val += root2.val;
root1.left = mergeTrees(root1.left, root2.left);
root1.right = mergeTrees(root1.right, root2.right);
return root1;
}
}
/**
* Time complexity : O(n). We traverse over a total of n nodes.
* Here, n refers to the smaller of the number of nodes in the two trees.
* Space complexity : O(n). The depth of stack can grow upto n in case of a skewed tree.
*/
class Solution {
public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
if(root1 == null && root2 == null) {
return null;
}
if(root1 == null) {
return root2;
}
if(root2 == null) {
return root1;
}
Stack<TreeNode[]> stack = new Stack();
stack.push(new TreeNode[]{root1, root2});
while(!stack.isEmpty()) {
TreeNode[] t = stack.pop();
if(t[1] == null) {
continue;
}
t[0].val += t[1].val;
if(t[0].left == null) {
t[0].left = t[1].left;
} else {
stack.push(new TreeNode[]{t[0].left, t[1].left});
}
if(t[0].right == null) {
t[0].right = t[1].right;
} else {
stack.push(new TreeNode[]{t[0].right, t[1].right});
}
}
return root1;
}
}
