236. Lowest Common Ancestor of a Binary Tree

发布时间：2020-12-14 04:47:26 所属栏目：大数据来源：网络整理

导读：问题描述： Given a binary tree,find the lowest common ancestor (LCA) of two given nodes in the tree. According to the?definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p?and q?as the lowest node in T

问题描述：

Given a binary tree,find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the?definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p?and q?as the lowest node in T that has both p?and q?as descendants (where we allow?a node to be a descendant of itself).”

Given the following binary tree:? root =?[3,5,1,6,2,8,null,7,4]

        _______3______
       /                  ___5__          ___1__
   /              /         6      _2       0       8
         /           7   4

Example 1:

Input: root = [3,4],p = 5,q = 1
Output: 3
Explanation: The LCA of of nodes  and  is 
513.

Example 2:

Input: root = [3,q = 4
Output: 5
Explanation: The LCA of nodes  and  is,since a node can be a descendant of itself
             according to the LCA definition.545

Note:

All of the nodes‘ values will be unique.
p and q are different and both values will?exist in the binary tree.

解题思路：

找两个节点最低的祖先节点，并且一个节点可以成为它自己的祖先节点。

我们可以先找到其中一个节点，这个时候我们需要遍历树，我在这里采用的是使用栈的前序遍历，当该节点值等于目标节点其中一个的值时，跳出循环。

找到的点为findN，未找到的为targetN

这个时候我们可以考虑两个点之间的关系：

　　1. targetN是findN的子节点

　　2.targetN不是findN的子节点

所以我们首先需要搜寻findN的子树，若包含targetN，则直接返回findN

若不包含，则需要从栈内弹出一个节点并且搜寻它的子树。

这里我们可以进行剪枝，因为此时栈顶是我们findN的父节点，并且我们已经搜索过了findN的子树，我们就不需要搜索它的左子树了。

但是targetN很可能是findN的父节点，所以我们先要对栈顶值进行检查，如果不是我们想要找的值，则检查它的右子树

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x),left(NULL),right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root,TreeNode* p,TreeNode* q) {
        stack<TreeNode*> stk;
        int restNode = 2;
        TreeNode* cur = root;
        while(cur || !stk.empty()){
            if(cur){
                if(cur->val == p->val || cur->val == q->val){
                    break;
                }
                stk.push(cur);
                cur = cur->left;
            }else{
                cur = stk.top();
                stk.pop();
                cur = cur->right;
            }
        }
        TreeNode* targetN = cur->val == p->val ? q : p;
        TreeNode* findN = cur;
        if(containsNode(cur,targetN)){
            return cur;
        }
        while(!stk.empty()){
            cur = stk.top();
            if(cur->val == targetN->val)
                break;
            stk.pop();
            if(containsNode(cur->right,targetN))
                break;
        }
        return cur;
    }
    bool containsNode(TreeNode* root,TreeNode* t){
        if(!root)
            return false;
        stack<TreeNode*> stk;
        TreeNode* cur = root;
        while(cur || !stk.empty()){
            if(cur){
                if(cur->val == t->val)
                    return true;
                stk.push(cur);
                cur = cur->left;
            }else{
                cur = stk.top();
                stk.pop();
                cur = cur->right;
            }
        }
        return false;
    }
};

（编辑：李大同）

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