05 August 2008
二叉树中第二小的节点
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
/*
https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/discuss/107158/Java-divide-and-conquer-solution
从左子树到右子树，如果子节点value与父节点value一样，就通过recursion找到下一个备选，否则就用当前子节点作为备选
*/
public int findSecondMinimumValue(TreeNode root) {
if (root == null) {
return -1;
}
if (root.left == null && root.right == null) {//只有一个root
return -1;
}
//最小的和第二小的
int left = root.left.val;
int right = root.right.val;
//if value same as root value, need to find the next candidate
if (root.left.val == root.val) {
left = findSecondMinimumValue(root.left);
}
if (root.right.val == root.val) {
right = findSecondMinimumValue(root.right);
}
if (left != -1 && right != -1) {
return Math.min(left, right);
} else if (left != -1) {
return left;
} else {
return right;
}
}
}