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package class30;

// 本题测试链接 : https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/

public class Code02_MinDepth {

// 不提交这个类

public static class TreeNode {

public int val;

public TreeNode left;

public TreeNode right;

public TreeNode(int x) {

val = x;

}

}

// 下面的方法是一般解

public static int minDepth1(TreeNode head) {

if (head == null) {

return 0;

}

return p(head);

}

// 返回x为头的树，最小深度是多少

public static int p(TreeNode x) {

if (x.left == null && x.right == null) {

return 1;

}

// 左右子树起码有一个不为空

int leftH = Integer.MAX_VALUE;

if (x.left != null) {

leftH = p(x.left);

}

int rightH = Integer.MAX_VALUE;

if (x.right != null) {

rightH = p(x.right);

}

return 1 + Math.min(leftH, rightH);

}

// 下面的方法是morris遍历的解

public static int minDepth2(TreeNode head) {

if (head == null) {

return 0;

}

TreeNode cur = head;

TreeNode mostRight = null;

int curLevel = 0;

int minHeight = Integer.MAX_VALUE;

while (cur != null) {

mostRight = cur.left;

if (mostRight != null) {

int rightBoardSize = 1;

while (mostRight.right != null && mostRight.right != cur) {

rightBoardSize++;

mostRight = mostRight.right;

}

if (mostRight.right == null) { // 第一次到达

curLevel++;

mostRight.right = cur;

cur = cur.left;

continue;

} else { // 第二次到达

if (mostRight.left == null) {

minHeight = Math.min(minHeight, curLevel);

}

curLevel -= rightBoardSize;

mostRight.right = null;

}

} else { // 只有一次到达

curLevel++;

}

cur = cur.right;

}

int finalRight = 1;

cur = head;

while (cur.right != null) {

finalRight++;

cur = cur.right;

}

if (cur.left == null && cur.right == null) {

minHeight = Math.min(minHeight, finalRight);

}

return minHeight;

}

}