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165 lines (158 loc) · 6.26 KB
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/*
* @lc app=leetcode id=117 lang=java
*
* [117] Populating Next Right Pointers in Each Node II
*/
// @lc code=start
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if (root == null) {
return root;
}
Queue<Node> nodeQueue = new LinkedList<>();
nodeQueue.add(root);
while (!nodeQueue.isEmpty()) {
int currLevelCnt = nodeQueue.size();
Node prevNode = null;
for (int i = 0; i < currLevelCnt; i++) {
Node currNode = nodeQueue.poll();
if (prevNode != null) {
prevNode.next = currNode;
}
if (currNode.left != null) {
nodeQueue.add(currNode.left);
}
if (currNode.right != null) {
nodeQueue.add(currNode.right);
}
prevNode = currNode;
}
}
return root;
}
// 这里由于子树有可能残缺，故需要平行扫描父节点同层的节点，找到他们的左右子节点
// recursive 方法
// public Node connect(Node root) {
// if (root == null) {
// return root;
// }
// if (root.left != null) {
// if (root.right != null) {
// root.left.next = root.right;
// }
// else {
// Node nextHasChildNode = root.next;
// if (nextHasChildNode != null) {
// while (nextHasChildNode.next != null && nextHasChildNode.left == null && nextHasChildNode.right == null) {
// nextHasChildNode = nextHasChildNode.next;
// }
// if (nextHasChildNode.left != null) {
// root.left.next = nextHasChildNode.left;
// }
// else if (nextHasChildNode.right != null) {
// root.left.next = nextHasChildNode.right;
// }
// }
// }
// }
// if (root.right != null) {
// Node nextHasChildNode = root.next;
// if (nextHasChildNode != null) {
// while (nextHasChildNode.next != null && nextHasChildNode.left == null && nextHasChildNode.right == null) {
// nextHasChildNode = nextHasChildNode.next;
// }
// if (nextHasChildNode.left != null) {
// root.right.next = nextHasChildNode.left;
// }
// else if (nextHasChildNode.right != null) {
// root.right.next = nextHasChildNode.right;
// }
// }
// }
// connect(root.right);
// connect(root.left);
// return root;
// }
// // 这里由于子树有可能残缺，故需要平行扫描父节点同层的节点，找到他们的左右子节点
// // constant space O(1)，runtime 接近O(N)
// public Node connect(Node root) {
// if (root == null) {
// return root;
// }
// Node currNode = null, startNode = root;
// while (startNode != null) {
// currNode = startNode;
// while (currNode != null) {
// if (currNode.left != null) {
// if (currNode.right != null) {
// currNode.left.next = currNode.right;
// }
// else {
// // 首先找到下一个有子节点的Node
// // 再把它的子节点和当前node的左子节点连起来
// Node nextHasChildNode = currNode.next;
// if (nextHasChildNode != null) {
// while (nextHasChildNode.next != null && nextHasChildNode.left == null && nextHasChildNode.right == null) {
// nextHasChildNode = nextHasChildNode.next;
// }
// if (nextHasChildNode.left != null) {
// currNode.left.next = nextHasChildNode.left;
// }
// else if (nextHasChildNode.right != null) {
// currNode.left.next = nextHasChildNode.right;
// }
// }
// }
// }
// if (currNode.right != null) {
// // 首先找到下一个有子节点的Node
// // 再把它的子节点和当前node的右子节点连起来
// Node nextHasChildNode = currNode.next;
// if (nextHasChildNode != null) {
// while (nextHasChildNode.next != null && nextHasChildNode.left == null && nextHasChildNode.right == null) {
// nextHasChildNode = nextHasChildNode.next;
// }
// if (nextHasChildNode.left != null) {
// currNode.right.next = nextHasChildNode.left;
// }
// else if (nextHasChildNode.right != null) {
// currNode.right.next = nextHasChildNode.right;
// }
// }
// }
// currNode = currNode.next;
// }
// // 找到下一个startNode，必须是有子节点并且和当前startNode在同一层的node
// while (startNode.next != null && startNode.left == null && startNode.right == null) {
// startNode = startNode.next;
// }
// if (startNode.left != null) {
// startNode = startNode.left;
// }
// else {
// startNode = startNode.right;
// }
// }
// return root;
// }
}
// @lc code=end