Problem
Given the root of a binary tree, return the leftmost value in the last row of the tree.
Example 1:
Input: root = [2,1,3]
Output: 1
Example 2:
Input: root = [1,2,3,4,null,5,6,null,null,7]
Output: 7
Constraints:
The number of nodes in the tree is in the range
[1, 10^4].-2^31 <= Node.val <= 2^31 - 1
Solution (Java)
/**
* 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;
* }
* }
*/
class Solution {
private int[] func(TreeNode root, int level) {
if (root.left == null && root.right == null) {
int[] a = new int[2];
a[0] = root.val;
a[1] = level;
return a;
}
int[] a = null;
int[] b = null;
if (root.left != null) {
a = func(root.left, level + 1);
}
if (root.right != null) {
b = func(root.right, level + 1);
}
if (a == null) {
return b;
} else if (b == null) {
return a;
} else {
if (a[1] >= b[1]) {
return a;
} else {
return b;
}
}
}
public int findBottomLeftValue(TreeNode root) {
int[] a = func(root, 0);
if (a != null && a.length > 0) {
return a[0];
}
return -1;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).